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Solving integrals/ (x^2+2x)lnx

Integral of (x^2+2x)lnx dx

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∫01(x2+2x)log⁡(x) dx\int\limits_{0}^{1} \left(x^{2} + 2 x\right) \log{\left(x \right)}\, dx 
Integral((x^2 + 2*x)*log(x), (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:

      ∫(ue3u+2ue2u) du\int \left(u e^{3 u} + 2 u e^{2 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)=uu{\left(u \right)} = u and let dv⁡(u)=e3u\operatorname{dv}{\left(u \right)} = e^{3 u}.

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

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

          ∫e3u3 du=∫e3u du3\int \frac{e^{3 u}}{3}\, du = \frac{\int e^{3 u}\, du}{3}

          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: e3u9\frac{e^{3 u}}{9}

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

          ∫2ue2u du=2∫ue2u du\int 2 u e^{2 u}\, du = 2 \int u 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)=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.

          2. 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: ue2u−e2u2u e^{2 u} - \frac{e^{2 u}}{2}

        The result is: ue3u3+ue2u−e3u9−e2u2\frac{u e^{3 u}}{3} + u e^{2 u} - \frac{e^{3 u}}{9} - \frac{e^{2 u}}{2}

      Now substitute uu back in:

      x3log⁡(x)3−x39+x2log⁡(x)−x22\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9} + x^{2} \log{\left(x \right)} - \frac{x^{2}}{2}

    Method #2

    1. Rewrite the integrand:

      (x2+2x)log⁡(x)=x2log⁡(x)+2xlog⁡(x)\left(x^{2} + 2 x\right) \log{\left(x \right)} = x^{2} \log{\left(x \right)} + 2 x \log{\left(x \right)}

    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:

        ∫ue3u du\int u 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)=uu{\left(u \right)} = u and let dv⁡(u)=e3u\operatorname{dv}{\left(u \right)} = e^{3 u}.

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

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

          ∫e3u3 du=∫e3u du3\int \frac{e^{3 u}}{3}\, du = \frac{\int e^{3 u}\, du}{3}

          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: e3u9\frac{e^{3 u}}{9}

        Now substitute uu back in:

        x3log⁡(x)3−x39\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9}

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

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

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

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

          ∫ue2u du\int u 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)=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.

          2. 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)2−x24\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}

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

      The result is: x3log⁡(x)3−x39+x2log⁡(x)−x22\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9} + x^{2} \log{\left(x \right)} - \frac{x^{2}}{2}

    Method #3

    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(x)=log⁡(x)u{\left(x \right)} = \log{\left(x \right)} and let dv⁡(x)=x2+2x\operatorname{dv}{\left(x \right)} = x^{2} + 2 x.

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

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

      1. Rewrite the integrand:

        x(x+2)=x2+2xx \left(x + 2\right) = x^{2} + 2 x

      2. Integrate term-by-term:

        1. The integral of xnx^{n} is xn+1n+1\frac{x^{n + 1}}{n + 1} when n≠−1n \neq -1:

          ∫x2 dx=x33\int x^{2}\, dx = \frac{x^{3}}{3}

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

          ∫2x dx=2∫x dx\int 2 x\, dx = 2 \int x\, dx

          1. The integral of xnx^{n} is xn+1n+1\frac{x^{n + 1}}{n + 1} when n≠−1n \neq -1:

            ∫x dx=x22\int x\, dx = \frac{x^{2}}{2}

          So, the result is: x2x^{2}

        The result is: x33+x2\frac{x^{3}}{3} + x^{2}

      Now evaluate the sub-integral.

    2. Rewrite the integrand:

      x33+x2x=x23+x\frac{\frac{x^{3}}{3} + x^{2}}{x} = \frac{x^{2}}{3} + x

    3. Integrate term-by-term:

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

        ∫x23 dx=∫x2 dx3\int \frac{x^{2}}{3}\, dx = \frac{\int x^{2}\, dx}{3}

        1. The integral of xnx^{n} is xn+1n+1\frac{x^{n + 1}}{n + 1} when n≠−1n \neq -1:

          ∫x2 dx=x33\int x^{2}\, dx = \frac{x^{3}}{3}

        So, the result is: x39\frac{x^{3}}{9}

      1. The integral of xnx^{n} is xn+1n+1\frac{x^{n + 1}}{n + 1} when n≠−1n \neq -1:

        ∫x dx=x22\int x\, dx = \frac{x^{2}}{2}

      The result is: x39+x22\frac{x^{3}}{9} + \frac{x^{2}}{2}

    Method #4

    1. Rewrite the integrand:

      (x2+2x)log⁡(x)=x2log⁡(x)+2xlog⁡(x)\left(x^{2} + 2 x\right) \log{\left(x \right)} = x^{2} \log{\left(x \right)} + 2 x \log{\left(x \right)}

    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:

        ∫ue3u du\int u 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)=uu{\left(u \right)} = u and let dv⁡(u)=e3u\operatorname{dv}{\left(u \right)} = e^{3 u}.

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

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

          ∫e3u3 du=∫e3u du3\int \frac{e^{3 u}}{3}\, du = \frac{\int e^{3 u}\, du}{3}

          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: e3u9\frac{e^{3 u}}{9}

        Now substitute uu back in:

        x3log⁡(x)3−x39\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9}

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

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

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

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

          ∫ue2u du\int u 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)=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.

          2. 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)2−x24\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}

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

      The result is: x3log⁡(x)3−x39+x2log⁡(x)−x22\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9} + x^{2} \log{\left(x \right)} - \frac{x^{2}}{2}

  2. Now simplify:

    x2(xlog⁡(x)3−x9+log⁡(x)−12)x^{2} \left(\frac{x \log{\left(x \right)}}{3} - \frac{x}{9} + \log{\left(x \right)} - \frac{1}{2}\right)

  3. Add the constant of integration:

    x2(xlog⁡(x)3−x9+log⁡(x)−12)+constantx^{2} \left(\frac{x \log{\left(x \right)}}{3} - \frac{x}{9} + \log{\left(x \right)} - \frac{1}{2}\right)+ \mathrm{constant}

The answer is:

x2(xlog⁡(x)3−x9+log⁡(x)−12)+constantx^{2} \left(\frac{x \log{\left(x \right)}}{3} - \frac{x}{9} + \log{\left(x \right)} - \frac{1}{2}\right)+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                          
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 | \x  + 2*x/*log(x) dx = C - -- - -- + x *log(x) + ---------
 |                            2    9                    3    
/
(x33+x2) log⁡x−2 x3+9 x218\left({{x^3}\over{3}}+x^2\right)\,\log x-{{2\,x^3+9\,x^2}\over{18}}
Simplify
The answer [src] 
-11 
----
 18
−1118-{{11}\over{18}} 
=
= 
-11 
----
 18
−1118- \frac{11}{18}
Simplify
Numerical answer [src] 
-0.611111111111111
-0.611111111111111

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

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