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Solving integrals/ (x+1)^2*ln^2(x)

Integral of (x+1)^2*ln^2(x) dx

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∫01(x+1)2log⁡(x)2 dx\int\limits_{0}^{1} \left(x + 1\right)^{2} \log{\left(x \right)}^{2}\, dx 
Integral((x + 1)^2*log(x)^2, (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=log⁡(x)u = \log{\left(x \right)}.

      Then let du=dxxdu = \frac{dx}{x} and substitute dudu:

      ∫(u2e3u+2u2e2u+u2eu) du\int \left(u^{2} e^{3 u} + 2 u^{2} e^{2 u} + u^{2} e^{u}\right)\, du

      1. Integrate term-by-term:

        1. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

          Let u(u)=u2u{\left(u \right)} = u^{2} and let dv⁡(u)=e3u\operatorname{dv}{\left(u \right)} = e^{3 u}.

          Then du⁡(u)=2u\operatorname{du}{\left(u \right)} = 2 u.

          To find v(u)v{\left(u \right)}:

          1. There are multiple ways to do this integral.

            Method #1

            1. Let u=3uu = 3 u.

              Then let du=3dudu = 3 du and substitute du3\frac{du}{3}:

              ∫eu9 du\int \frac{e^{u}}{9}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                ∫eu3 du=∫eu du3\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}

                1. The integral of the exponential function is itself.

                  ∫eu du=eu\int e^{u}\, du = e^{u}

                So, the result is: eu3\frac{e^{u}}{3}

              Now substitute uu back in:

              e3u3\frac{e^{3 u}}{3}

            Method #2

            1. Let u=e3uu = e^{3 u}.

              Then let du=3e3ududu = 3 e^{3 u} du and substitute du3\frac{du}{3}:

              ∫19 du\int \frac{1}{9}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                ∫13 du=∫1 du3\int \frac{1}{3}\, du = \frac{\int 1\, du}{3}

                1. The integral of a constant is the constant times the variable of integration:

                  ∫1 du=u\int 1\, du = u

                So, the result is: u3\frac{u}{3}

              Now substitute uu back in:

              e3u3\frac{e^{3 u}}{3}

          Now evaluate the sub-integral.

        2. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

          Let u(u)=2u3u{\left(u \right)} = \frac{2 u}{3} and let dv⁡(u)=e3u\operatorname{dv}{\left(u \right)} = e^{3 u}.

          Then du⁡(u)=23\operatorname{du}{\left(u \right)} = \frac{2}{3}.

          To find v(u)v{\left(u \right)}:

          1. Let u=3uu = 3 u.

            Then let du=3dudu = 3 du and substitute du3\frac{du}{3}:

            ∫eu9 du\int \frac{e^{u}}{9}\, du

            1. The integral of a constant times a function is the constant times the integral of the function:

              ∫eu3 du=∫eu du3\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}

              1. The integral of the exponential function is itself.

                ∫eu du=eu\int e^{u}\, du = e^{u}

              So, the result is: eu3\frac{e^{u}}{3}

            Now substitute uu back in:

            e3u3\frac{e^{3 u}}{3}

          Now evaluate the sub-integral.

        3. The integral of a constant times a function is the constant times the integral of the function:

          ∫2e3u9 du=2∫e3u du9\int \frac{2 e^{3 u}}{9}\, du = \frac{2 \int e^{3 u}\, du}{9}

          1. Let u=3uu = 3 u.

            Then let du=3dudu = 3 du and substitute du3\frac{du}{3}:

            ∫eu9 du\int \frac{e^{u}}{9}\, du

            1. The integral of a constant times a function is the constant times the integral of the function:

              ∫eu3 du=∫eu du3\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}

              1. The integral of the exponential function is itself.

                ∫eu du=eu\int e^{u}\, du = e^{u}

              So, the result is: eu3\frac{e^{u}}{3}

            Now substitute uu back in:

            e3u3\frac{e^{3 u}}{3}

          So, the result is: 2e3u27\frac{2 e^{3 u}}{27}

        1. The integral of a constant times a function is the constant times the integral of the function:

          ∫2u2e2u du=2∫u2e2u du\int 2 u^{2} e^{2 u}\, du = 2 \int u^{2} e^{2 u}\, du

          1. Use integration by parts:

            ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

            Let u(u)=u2u{\left(u \right)} = u^{2} and let dv⁡(u)=e2u\operatorname{dv}{\left(u \right)} = e^{2 u}.

            Then du⁡(u)=2u\operatorname{du}{\left(u \right)} = 2 u.

            To find v(u)v{\left(u \right)}:

            1. Let u=2uu = 2 u.

              Then let du=2dudu = 2 du and substitute du2\frac{du}{2}:

              ∫eu4 du\int \frac{e^{u}}{4}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                ∫eu2 du=∫eu du2\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}

                1. The integral of the exponential function is itself.

                  ∫eu du=eu\int e^{u}\, du = e^{u}

                So, the result is: eu2\frac{e^{u}}{2}

              Now substitute uu back in:

              e2u2\frac{e^{2 u}}{2}

            Now evaluate the sub-integral.

          2. Use integration by parts:

            ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

            Let u(u)=uu{\left(u \right)} = u and let dv⁡(u)=e2u\operatorname{dv}{\left(u \right)} = e^{2 u}.

            Then du⁡(u)=1\operatorname{du}{\left(u \right)} = 1.

            To find v(u)v{\left(u \right)}:

            1. Let u=2uu = 2 u.

              Then let du=2dudu = 2 du and substitute du2\frac{du}{2}:

              ∫eu4 du\int \frac{e^{u}}{4}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                ∫eu2 du=∫eu du2\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}

                1. The integral of the exponential function is itself.

                  ∫eu du=eu\int e^{u}\, du = e^{u}

                So, the result is: eu2\frac{e^{u}}{2}

              Now substitute uu back in:

              e2u2\frac{e^{2 u}}{2}

            Now evaluate the sub-integral.

          3. The integral of a constant times a function is the constant times the integral of the function:

            ∫e2u2 du=∫e2u du2\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}

            1. Let u=2uu = 2 u.

              Then let du=2dudu = 2 du and substitute du2\frac{du}{2}:

              ∫eu4 du\int \frac{e^{u}}{4}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                ∫eu2 du=∫eu du2\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}

                1. The integral of the exponential function is itself.

                  ∫eu du=eu\int e^{u}\, du = e^{u}

                So, the result is: eu2\frac{e^{u}}{2}

              Now substitute uu back in:

              e2u2\frac{e^{2 u}}{2}

            So, the result is: e2u4\frac{e^{2 u}}{4}

          So, the result is: u2e2u−ue2u+e2u2u^{2} e^{2 u} - u e^{2 u} + \frac{e^{2 u}}{2}

        1. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

          Let u(u)=u2u{\left(u \right)} = u^{2} and let dv⁡(u)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

          Then du⁡(u)=2u\operatorname{du}{\left(u \right)} = 2 u.

          To find v(u)v{\left(u \right)}:

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          Now evaluate the sub-integral.

        2. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

          Let u(u)=2uu{\left(u \right)} = 2 u and let dv⁡(u)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

          Then du⁡(u)=2\operatorname{du}{\left(u \right)} = 2.

          To find v(u)v{\left(u \right)}:

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          Now evaluate the sub-integral.

        3. The integral of a constant times a function is the constant times the integral of the function:

          ∫2eu du=2∫eu du\int 2 e^{u}\, du = 2 \int e^{u}\, du

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          So, the result is: 2eu2 e^{u}

        The result is: u2e3u3+u2e2u+u2eu−2ue3u9−ue2u−2ueu+2e3u27+e2u2+2eu\frac{u^{2} e^{3 u}}{3} + u^{2} e^{2 u} + u^{2} e^{u} - \frac{2 u e^{3 u}}{9} - u e^{2 u} - 2 u e^{u} + \frac{2 e^{3 u}}{27} + \frac{e^{2 u}}{2} + 2 e^{u}

      Now substitute uu back in:

      x3log⁡(x)23−2x3log⁡(x)9+2x327+x2log⁡(x)2−x2log⁡(x)+x22+xlog⁡(x)2−2xlog⁡(x)+2x\frac{x^{3} \log{\left(x \right)}^{2}}{3} - \frac{2 x^{3} \log{\left(x \right)}}{9} + \frac{2 x^{3}}{27} + x^{2} \log{\left(x \right)}^{2} - x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + x \log{\left(x \right)}^{2} - 2 x \log{\left(x \right)} + 2 x

    Method #2

    1. Rewrite the integrand:

      (x+1)2log⁡(x)2=x2log⁡(x)2+2xlog⁡(x)2+log⁡(x)2\left(x + 1\right)^{2} \log{\left(x \right)}^{2} = x^{2} \log{\left(x \right)}^{2} + 2 x \log{\left(x \right)}^{2} + \log{\left(x \right)}^{2}

    2. Integrate term-by-term:

      1. Let u=log⁡(x)u = \log{\left(x \right)}.

        Then let du=dxxdu = \frac{dx}{x} and substitute dudu:

        ∫u2e3u du\int u^{2} e^{3 u}\, du

        1. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

          Let u(u)=u2u{\left(u \right)} = u^{2} and let dv⁡(u)=e3u\operatorname{dv}{\left(u \right)} = e^{3 u}.

          Then du⁡(u)=2u\operatorname{du}{\left(u \right)} = 2 u.

          To find v(u)v{\left(u \right)}:

          1. Let u=3uu = 3 u.

            Then let du=3dudu = 3 du and substitute du3\frac{du}{3}:

            ∫eu9 du\int \frac{e^{u}}{9}\, du

            1. The integral of a constant times a function is the constant times the integral of the function:

              ∫eu3 du=∫eu du3\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}

              1. The integral of the exponential function is itself.

                ∫eu du=eu\int e^{u}\, du = e^{u}

              So, the result is: eu3\frac{e^{u}}{3}

            Now substitute uu back in:

            e3u3\frac{e^{3 u}}{3}

          Now evaluate the sub-integral.

        2. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

          Let u(u)=2u3u{\left(u \right)} = \frac{2 u}{3} and let dv⁡(u)=e3u\operatorname{dv}{\left(u \right)} = e^{3 u}.

          Then du⁡(u)=23\operatorname{du}{\left(u \right)} = \frac{2}{3}.

          To find v(u)v{\left(u \right)}:

          1. Let u=3uu = 3 u.

            Then let du=3dudu = 3 du and substitute du3\frac{du}{3}:

            ∫eu9 du\int \frac{e^{u}}{9}\, du

            1. The integral of a constant times a function is the constant times the integral of the function:

              ∫eu3 du=∫eu du3\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}

              1. The integral of the exponential function is itself.

                ∫eu du=eu\int e^{u}\, du = e^{u}

              So, the result is: eu3\frac{e^{u}}{3}

            Now substitute uu back in:

            e3u3\frac{e^{3 u}}{3}

          Now evaluate the sub-integral.

        3. The integral of a constant times a function is the constant times the integral of the function:

          ∫2e3u9 du=2∫e3u du9\int \frac{2 e^{3 u}}{9}\, du = \frac{2 \int e^{3 u}\, du}{9}

          1. Let u=3uu = 3 u.

            Then let du=3dudu = 3 du and substitute du3\frac{du}{3}:

            ∫eu9 du\int \frac{e^{u}}{9}\, du

            1. The integral of a constant times a function is the constant times the integral of the function:

              ∫eu3 du=∫eu du3\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}

              1. The integral of the exponential function is itself.

                ∫eu du=eu\int e^{u}\, du = e^{u}

              So, the result is: eu3\frac{e^{u}}{3}

            Now substitute uu back in:

            e3u3\frac{e^{3 u}}{3}

          So, the result is: 2e3u27\frac{2 e^{3 u}}{27}

        Now substitute uu back in:

        x3log⁡(x)23−2x3log⁡(x)9+2x327\frac{x^{3} \log{\left(x \right)}^{2}}{3} - \frac{2 x^{3} \log{\left(x \right)}}{9} + \frac{2 x^{3}}{27}

      1. The integral of a constant times a function is the constant times the integral of the function:

        ∫2xlog⁡(x)2 dx=2∫xlog⁡(x)2 dx\int 2 x \log{\left(x \right)}^{2}\, dx = 2 \int x \log{\left(x \right)}^{2}\, dx

        1. Let u=log⁡(x)u = \log{\left(x \right)}.

          Then let du=dxxdu = \frac{dx}{x} and substitute dudu:

          ∫u2e2u du\int u^{2} e^{2 u}\, du

          1. Use integration by parts:

            ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

            Let u(u)=u2u{\left(u \right)} = u^{2} and let dv⁡(u)=e2u\operatorname{dv}{\left(u \right)} = e^{2 u}.

            Then du⁡(u)=2u\operatorname{du}{\left(u \right)} = 2 u.

            To find v(u)v{\left(u \right)}:

            1. Let u=2uu = 2 u.

              Then let du=2dudu = 2 du and substitute du2\frac{du}{2}:

              ∫eu4 du\int \frac{e^{u}}{4}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                ∫eu2 du=∫eu du2\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}

                1. The integral of the exponential function is itself.

                  ∫eu du=eu\int e^{u}\, du = e^{u}

                So, the result is: eu2\frac{e^{u}}{2}

              Now substitute uu back in:

              e2u2\frac{e^{2 u}}{2}

            Now evaluate the sub-integral.

          2. Use integration by parts:

            ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

            Let u(u)=uu{\left(u \right)} = u and let dv⁡(u)=e2u\operatorname{dv}{\left(u \right)} = e^{2 u}.

            Then du⁡(u)=1\operatorname{du}{\left(u \right)} = 1.

            To find v(u)v{\left(u \right)}:

            1. Let u=2uu = 2 u.

              Then let du=2dudu = 2 du and substitute du2\frac{du}{2}:

              ∫eu4 du\int \frac{e^{u}}{4}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                ∫eu2 du=∫eu du2\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}

                1. The integral of the exponential function is itself.

                  ∫eu du=eu\int e^{u}\, du = e^{u}

                So, the result is: eu2\frac{e^{u}}{2}

              Now substitute uu back in:

              e2u2\frac{e^{2 u}}{2}

            Now evaluate the sub-integral.

          3. The integral of a constant times a function is the constant times the integral of the function:

            ∫e2u2 du=∫e2u du2\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}

            1. Let u=2uu = 2 u.

              Then let du=2dudu = 2 du and substitute du2\frac{du}{2}:

              ∫eu4 du\int \frac{e^{u}}{4}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                ∫eu2 du=∫eu du2\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}

                1. The integral of the exponential function is itself.

                  ∫eu du=eu\int e^{u}\, du = e^{u}

                So, the result is: eu2\frac{e^{u}}{2}

              Now substitute uu back in:

              e2u2\frac{e^{2 u}}{2}

            So, the result is: e2u4\frac{e^{2 u}}{4}

          Now substitute uu back in:

          x2log⁡(x)22−x2log⁡(x)2+x24\frac{x^{2} \log{\left(x \right)}^{2}}{2} - \frac{x^{2} \log{\left(x \right)}}{2} + \frac{x^{2}}{4}

        So, the result is: x2log⁡(x)2−x2log⁡(x)+x22x^{2} \log{\left(x \right)}^{2} - x^{2} \log{\left(x \right)} + \frac{x^{2}}{2}

      1. Let u=log⁡(x)u = \log{\left(x \right)}.

        Then let du=dxxdu = \frac{dx}{x} and substitute dudu:

        ∫u2eu du\int u^{2} e^{u}\, du

        1. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

          Let u(u)=u2u{\left(u \right)} = u^{2} and let dv⁡(u)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

          Then du⁡(u)=2u\operatorname{du}{\left(u \right)} = 2 u.

          To find v(u)v{\left(u \right)}:

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          Now evaluate the sub-integral.

        2. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

          Let u(u)=2uu{\left(u \right)} = 2 u and let dv⁡(u)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

          Then du⁡(u)=2\operatorname{du}{\left(u \right)} = 2.

          To find v(u)v{\left(u \right)}:

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          Now evaluate the sub-integral.

        3. The integral of a constant times a function is the constant times the integral of the function:

          ∫2eu du=2∫eu du\int 2 e^{u}\, du = 2 \int e^{u}\, du

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          So, the result is: 2eu2 e^{u}

        Now substitute uu back in:

        xlog⁡(x)2−2xlog⁡(x)+2xx \log{\left(x \right)}^{2} - 2 x \log{\left(x \right)} + 2 x

      The result is: x3log⁡(x)23−2x3log⁡(x)9+2x327+x2log⁡(x)2−x2log⁡(x)+x22+xlog⁡(x)2−2xlog⁡(x)+2x\frac{x^{3} \log{\left(x \right)}^{2}}{3} - \frac{2 x^{3} \log{\left(x \right)}}{9} + \frac{2 x^{3}}{27} + x^{2} \log{\left(x \right)}^{2} - x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + x \log{\left(x \right)}^{2} - 2 x \log{\left(x \right)} + 2 x

    Method #3

    1. Rewrite the integrand:

      (x+1)2log⁡(x)2=x2log⁡(x)2+2xlog⁡(x)2+log⁡(x)2\left(x + 1\right)^{2} \log{\left(x \right)}^{2} = x^{2} \log{\left(x \right)}^{2} + 2 x \log{\left(x \right)}^{2} + \log{\left(x \right)}^{2}

    2. Integrate term-by-term:

      1. Let u=log⁡(x)u = \log{\left(x \right)}.

        Then let du=dxxdu = \frac{dx}{x} and substitute dudu:

        ∫u2e3u du\int u^{2} e^{3 u}\, du

        1. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

          Let u(u)=u2u{\left(u \right)} = u^{2} and let dv⁡(u)=e3u\operatorname{dv}{\left(u \right)} = e^{3 u}.

          Then du⁡(u)=2u\operatorname{du}{\left(u \right)} = 2 u.

          To find v(u)v{\left(u \right)}:

          1. Let u=3uu = 3 u.

            Then let du=3dudu = 3 du and substitute du3\frac{du}{3}:

            ∫eu9 du\int \frac{e^{u}}{9}\, du

            1. The integral of a constant times a function is the constant times the integral of the function:

              ∫eu3 du=∫eu du3\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}

              1. The integral of the exponential function is itself.

                ∫eu du=eu\int e^{u}\, du = e^{u}

              So, the result is: eu3\frac{e^{u}}{3}

            Now substitute uu back in:

            e3u3\frac{e^{3 u}}{3}

          Now evaluate the sub-integral.

        2. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

          Let u(u)=2u3u{\left(u \right)} = \frac{2 u}{3} and let dv⁡(u)=e3u\operatorname{dv}{\left(u \right)} = e^{3 u}.

          Then du⁡(u)=23\operatorname{du}{\left(u \right)} = \frac{2}{3}.

          To find v(u)v{\left(u \right)}:

          1. Let u=3uu = 3 u.

            Then let du=3dudu = 3 du and substitute du3\frac{du}{3}:

            ∫eu9 du\int \frac{e^{u}}{9}\, du

            1. The integral of a constant times a function is the constant times the integral of the function:

              ∫eu3 du=∫eu du3\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}

              1. The integral of the exponential function is itself.

                ∫eu du=eu\int e^{u}\, du = e^{u}

              So, the result is: eu3\frac{e^{u}}{3}

            Now substitute uu back in:

            e3u3\frac{e^{3 u}}{3}

          Now evaluate the sub-integral.

        3. The integral of a constant times a function is the constant times the integral of the function:

          ∫2e3u9 du=2∫e3u du9\int \frac{2 e^{3 u}}{9}\, du = \frac{2 \int e^{3 u}\, du}{9}

          1. Let u=3uu = 3 u.

            Then let du=3dudu = 3 du and substitute du3\frac{du}{3}:

            ∫eu9 du\int \frac{e^{u}}{9}\, du

            1. The integral of a constant times a function is the constant times the integral of the function:

              ∫eu3 du=∫eu du3\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}

              1. The integral of the exponential function is itself.

                ∫eu du=eu\int e^{u}\, du = e^{u}

              So, the result is: eu3\frac{e^{u}}{3}

            Now substitute uu back in:

            e3u3\frac{e^{3 u}}{3}

          So, the result is: 2e3u27\frac{2 e^{3 u}}{27}

        Now substitute uu back in:

        x3log⁡(x)23−2x3log⁡(x)9+2x327\frac{x^{3} \log{\left(x \right)}^{2}}{3} - \frac{2 x^{3} \log{\left(x \right)}}{9} + \frac{2 x^{3}}{27}

      1. The integral of a constant times a function is the constant times the integral of the function:

        ∫2xlog⁡(x)2 dx=2∫xlog⁡(x)2 dx\int 2 x \log{\left(x \right)}^{2}\, dx = 2 \int x \log{\left(x \right)}^{2}\, dx

        1. Let u=log⁡(x)u = \log{\left(x \right)}.

          Then let du=dxxdu = \frac{dx}{x} and substitute dudu:

          ∫u2e2u du\int u^{2} e^{2 u}\, du

          1. Use integration by parts:

            ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

            Let u(u)=u2u{\left(u \right)} = u^{2} and let dv⁡(u)=e2u\operatorname{dv}{\left(u \right)} = e^{2 u}.

            Then du⁡(u)=2u\operatorname{du}{\left(u \right)} = 2 u.

            To find v(u)v{\left(u \right)}:

            1. Let u=2uu = 2 u.

              Then let du=2dudu = 2 du and substitute du2\frac{du}{2}:

              ∫eu4 du\int \frac{e^{u}}{4}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                ∫eu2 du=∫eu du2\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}

                1. The integral of the exponential function is itself.

                  ∫eu du=eu\int e^{u}\, du = e^{u}

                So, the result is: eu2\frac{e^{u}}{2}

              Now substitute uu back in:

              e2u2\frac{e^{2 u}}{2}

            Now evaluate the sub-integral.

          2. Use integration by parts:

            ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

            Let u(u)=uu{\left(u \right)} = u and let dv⁡(u)=e2u\operatorname{dv}{\left(u \right)} = e^{2 u}.

            Then du⁡(u)=1\operatorname{du}{\left(u \right)} = 1.

            To find v(u)v{\left(u \right)}:

            1. Let u=2uu = 2 u.

              Then let du=2dudu = 2 du and substitute du2\frac{du}{2}:

              ∫eu4 du\int \frac{e^{u}}{4}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                ∫eu2 du=∫eu du2\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}

                1. The integral of the exponential function is itself.

                  ∫eu du=eu\int e^{u}\, du = e^{u}

                So, the result is: eu2\frac{e^{u}}{2}

              Now substitute uu back in:

              e2u2\frac{e^{2 u}}{2}

            Now evaluate the sub-integral.

          3. The integral of a constant times a function is the constant times the integral of the function:

            ∫e2u2 du=∫e2u du2\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}

            1. Let u=2uu = 2 u.

              Then let du=2dudu = 2 du and substitute du2\frac{du}{2}:

              ∫eu4 du\int \frac{e^{u}}{4}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                ∫eu2 du=∫eu du2\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}

                1. The integral of the exponential function is itself.

                  ∫eu du=eu\int e^{u}\, du = e^{u}

                So, the result is: eu2\frac{e^{u}}{2}

              Now substitute uu back in:

              e2u2\frac{e^{2 u}}{2}

            So, the result is: e2u4\frac{e^{2 u}}{4}

          Now substitute uu back in:

          x2log⁡(x)22−x2log⁡(x)2+x24\frac{x^{2} \log{\left(x \right)}^{2}}{2} - \frac{x^{2} \log{\left(x \right)}}{2} + \frac{x^{2}}{4}

        So, the result is: x2log⁡(x)2−x2log⁡(x)+x22x^{2} \log{\left(x \right)}^{2} - x^{2} \log{\left(x \right)} + \frac{x^{2}}{2}

      1. Let u=log⁡(x)u = \log{\left(x \right)}.

        Then let du=dxxdu = \frac{dx}{x} and substitute dudu:

        ∫u2eu du\int u^{2} e^{u}\, du

        1. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

          Let u(u)=u2u{\left(u \right)} = u^{2} and let dv⁡(u)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

          Then du⁡(u)=2u\operatorname{du}{\left(u \right)} = 2 u.

          To find v(u)v{\left(u \right)}:

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          Now evaluate the sub-integral.

        2. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

          Let u(u)=2uu{\left(u \right)} = 2 u and let dv⁡(u)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

          Then du⁡(u)=2\operatorname{du}{\left(u \right)} = 2.

          To find v(u)v{\left(u \right)}:

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          Now evaluate the sub-integral.

        3. The integral of a constant times a function is the constant times the integral of the function:

          ∫2eu du=2∫eu du\int 2 e^{u}\, du = 2 \int e^{u}\, du

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          So, the result is: 2eu2 e^{u}

        Now substitute uu back in:

        xlog⁡(x)2−2xlog⁡(x)+2xx \log{\left(x \right)}^{2} - 2 x \log{\left(x \right)} + 2 x

      The result is: x3log⁡(x)23−2x3log⁡(x)9+2x327+x2log⁡(x)2−x2log⁡(x)+x22+xlog⁡(x)2−2xlog⁡(x)+2x\frac{x^{3} \log{\left(x \right)}^{2}}{3} - \frac{2 x^{3} \log{\left(x \right)}}{9} + \frac{2 x^{3}}{27} + x^{2} \log{\left(x \right)}^{2} - x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + x \log{\left(x \right)}^{2} - 2 x \log{\left(x \right)} + 2 x

  2. Now simplify:

    x(18x2log⁡(x)2−12x2log⁡(x)+4x2+54xlog⁡(x)2−54xlog⁡(x)+27x+54log⁡(x)2−108log⁡(x)+108)54\frac{x \left(18 x^{2} \log{\left(x \right)}^{2} - 12 x^{2} \log{\left(x \right)} + 4 x^{2} + 54 x \log{\left(x \right)}^{2} - 54 x \log{\left(x \right)} + 27 x + 54 \log{\left(x \right)}^{2} - 108 \log{\left(x \right)} + 108\right)}{54}

  3. Add the constant of integration:

    x(18x2log⁡(x)2−12x2log⁡(x)+4x2+54xlog⁡(x)2−54xlog⁡(x)+27x+54log⁡(x)2−108log⁡(x)+108)54+constant\frac{x \left(18 x^{2} \log{\left(x \right)}^{2} - 12 x^{2} \log{\left(x \right)} + 4 x^{2} + 54 x \log{\left(x \right)}^{2} - 54 x \log{\left(x \right)} + 27 x + 54 \log{\left(x \right)}^{2} - 108 \log{\left(x \right)} + 108\right)}{54}+ \mathrm{constant}

The answer is:

x(18x2log⁡(x)2−12x2log⁡(x)+4x2+54xlog⁡(x)2−54xlog⁡(x)+27x+54log⁡(x)2−108log⁡(x)+108)54+constant\frac{x \left(18 x^{2} \log{\left(x \right)}^{2} - 12 x^{2} \log{\left(x \right)} + 4 x^{2} + 54 x \log{\left(x \right)}^{2} - 54 x \log{\left(x \right)} + 27 x + 54 \log{\left(x \right)}^{2} - 108 \log{\left(x \right)} + 108\right)}{54}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                                                      
 |                            2            3                                                        3           3    2   
 |        2    2             x          2*x         2       2    2       2                       2*x *log(x)   x *log (x)
 | (x + 1) *log (x) dx = C + -- + 2*x + ---- + x*log (x) + x *log (x) - x *log(x) - 2*x*log(x) - ----------- + ----------
 |                           2           27                                                           9            3     
/
x3 (9 (log⁡x)2−6 log⁡x+2)27+x2 (2 (log⁡x)2−2 log⁡x+1)2+x ((log⁡x)2−2 log⁡x+2){{x^3\,\left(9\,\left(\log x\right)^2-6\,\log x+2\right)}\over{27}} +{{x^2\,\left(2\,\left(\log x\right)^2-2\,\log x+1\right)}\over{2}}+ x\,\left(\left(\log x\right)^2-2\,\log x+2\right)
Simplify
The answer [src] 
139
---
 54
13954{{139}\over{54}} 
=
= 
139
---
 54
13954\frac{139}{54}
Упростить
Numerical answer [src] 
2.57407407407407
2.57407407407407

    Use the examples entering the upper and lower limits of integration.

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