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Solving integrals/ x(1-ln(x))²+1

Integral of x(1-ln(x))²+1 dx

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The solution

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∫13(x(1−log⁡(x))2+1) dx\int\limits_{1}^{3} \left(x \left(1 - \log{\left(x \right)}\right)^{2} + 1\right)\, dx 
Integral(x*(1 - log(x))^2 + 1, (x, 1, 3))
Detail solution

  1. Integrate term-by-term:

    1. There are multiple ways to do this integral.

      Method #1

      1. Rewrite the integrand:

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

      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:

          ∫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}:

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

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

                False\text{False}

                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}:

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

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

                False\text{False}

                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}:

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

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

                False\text{False}

                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}

        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 \left(- 2 x \log{\left(x \right)}\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}:

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

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

                  False\text{False}

                  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}:

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

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

                  False\text{False}

                  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)+x22- x^{2} \log{\left(x \right)} + \frac{x^{2}}{2}

        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: x2log⁡(x)22−3x2log⁡(x)2+5x24\frac{x^{2} \log{\left(x \right)}^{2}}{2} - \frac{3 x^{2} \log{\left(x \right)}}{2} + \frac{5 x^{2}}{4}

      Method #2

      1. Rewrite the integrand:

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

      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:

          ∫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}:

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

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

                False\text{False}

                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}:

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

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

                False\text{False}

                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}:

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

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

                False\text{False}

                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}

        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 \left(- 2 x \log{\left(x \right)}\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}:

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

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

                  False\text{False}

                  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}:

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

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

                  False\text{False}

                  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)+x22- x^{2} \log{\left(x \right)} + \frac{x^{2}}{2}

        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: x2log⁡(x)22−3x2log⁡(x)2+5x24\frac{x^{2} \log{\left(x \right)}^{2}}{2} - \frac{3 x^{2} \log{\left(x \right)}}{2} + \frac{5 x^{2}}{4}

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

      ∫1 dx=x\int 1\, dx = x

    The result is: x2log⁡(x)22−3x2log⁡(x)2+5x24+x\frac{x^{2} \log{\left(x \right)}^{2}}{2} - \frac{3 x^{2} \log{\left(x \right)}}{2} + \frac{5 x^{2}}{4} + x

  2. Now simplify:

    x(2xlog⁡(x)2−6xlog⁡(x)+5x+4)4\frac{x \left(2 x \log{\left(x \right)}^{2} - 6 x \log{\left(x \right)} + 5 x + 4\right)}{4}

  3. Add the constant of integration:

    x(2xlog⁡(x)2−6xlog⁡(x)+5x+4)4+constant\frac{x \left(2 x \log{\left(x \right)}^{2} - 6 x \log{\left(x \right)} + 5 x + 4\right)}{4}+ \mathrm{constant}

The answer is:

x(2xlog⁡(x)2−6xlog⁡(x)+5x+4)4+constant\frac{x \left(2 x \log{\left(x \right)}^{2} - 6 x \log{\left(x \right)} + 5 x + 4\right)}{4}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                  
 |                                       2    2    2         2       
 | /              2    \              5*x    x *log (x)   3*x *log(x)
 | \x*(1 - log(x))  + 1/ dx = C + x + ---- + ---------- - -----------
 |                                     4         2             2     
/
∫(x(1−log⁡(x))2+1) dx=C+x2log⁡(x)22−3x2log⁡(x)2+5x24+x\int \left(x \left(1 - \log{\left(x \right)}\right)^{2} + 1\right)\, dx = C + \frac{x^{2} \log{\left(x \right)}^{2}}{2} - \frac{3 x^{2} \log{\left(x \right)}}{2} + \frac{5 x^{2}}{4} + x
Simplify
The graph
1.03.01.21.41.61.82.02.22.42.62.805
Plot the graph f(x)
The answer [src] 
                      2   
     27*log(3)   9*log (3)
12 - --------- + ---------
         2           2
−27log⁡(3)2+9log⁡(3)22+12- \frac{27 \log{\left(3 \right)}}{2} + \frac{9 \log{\left(3 \right)}^{2}}{2} + 12 
=
= 
                      2   
     27*log(3)   9*log (3)
12 - --------- + ---------
         2           2
−27log⁡(3)2+9log⁡(3)22+12- \frac{27 \log{\left(3 \right)}}{2} + \frac{9 \log{\left(3 \right)}^{2}}{2} + 12 
12 - 27*log(3)/2 + 9*log(3)^2/2
Numerical answer [src] 
2.60000442663714
2.60000442663714

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

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