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Solving integrals/ x-1/x(ln(x)-x)^2

Integral of x-1/x(ln(x)-x)^2 dx

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

  1. 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:

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

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

      ∫(−(−x+log⁡(x))2x) dx=−∫(−x+log⁡(x))2x dx\int \left(- \frac{\left(- x + \log{\left(x \right)}\right)^{2}}{x}\right)\, dx = - \int \frac{\left(- x + \log{\left(x \right)}\right)^{2}}{x}\, dx

      1. There are multiple ways to do this integral.

        Method #1

        1. Let u=1xu = \frac{1}{x}.

          Then let du=−dxx2du = - \frac{dx}{x^{2}} and substitute −du- du:

          ∫(−u2log⁡(1u)2−2ulog⁡(1u)+1u3) du\int \left(- \frac{u^{2} \log{\left(\frac{1}{u} \right)}^{2} - 2 u \log{\left(\frac{1}{u} \right)} + 1}{u^{3}}\right)\, du

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

            ∫u2log⁡(1u)2−2ulog⁡(1u)+1u3 du=−∫u2log⁡(1u)2−2ulog⁡(1u)+1u3 du\int \frac{u^{2} \log{\left(\frac{1}{u} \right)}^{2} - 2 u \log{\left(\frac{1}{u} \right)} + 1}{u^{3}}\, du = - \int \frac{u^{2} \log{\left(\frac{1}{u} \right)}^{2} - 2 u \log{\left(\frac{1}{u} \right)} + 1}{u^{3}}\, du

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

              Then let du=−duudu = - \frac{du}{u} and substitute dudu:

              ∫(−u2+2ueu−e2u) du\int \left(- u^{2} + 2 u e^{u} - e^{2 u}\right)\, du

              1. Integrate term-by-term:

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

                  ∫(−u2) du=−∫u2 du\int \left(- u^{2}\right)\, du = - \int u^{2}\, du

                  1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                    ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

                  So, the result is: −u33- \frac{u^{3}}{3}

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

                  ∫2ueu du=2∫ueu du\int 2 u e^{u}\, du = 2 \int u 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)=uu{\left(u \right)} = u and let dv⁡(u)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

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

                    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. The integral of the exponential function is itself.

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

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

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

                  ∫(−e2u) du=−∫e2u du\int \left(- e^{2 u}\right)\, du = - \int e^{2 u}\, du

                  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: −e2u2- \frac{e^{2 u}}{2}

                The result is: −u33+2ueu−e2u2−2eu- \frac{u^{3}}{3} + 2 u e^{u} - \frac{e^{2 u}}{2} - 2 e^{u}

              Now substitute uu back in:

              −log⁡(1u)33+2log⁡(1u)u−2u−12u2- \frac{\log{\left(\frac{1}{u} \right)}^{3}}{3} + \frac{2 \log{\left(\frac{1}{u} \right)}}{u} - \frac{2}{u} - \frac{1}{2 u^{2}}

            So, the result is: log⁡(1u)33−2log⁡(1u)u+2u+12u2\frac{\log{\left(\frac{1}{u} \right)}^{3}}{3} - \frac{2 \log{\left(\frac{1}{u} \right)}}{u} + \frac{2}{u} + \frac{1}{2 u^{2}}

          Now substitute uu back in:

          x22−2xlog⁡(x)+2x+log⁡(x)33\frac{x^{2}}{2} - 2 x \log{\left(x \right)} + 2 x + \frac{\log{\left(x \right)}^{3}}{3}

        Method #2

        1. Rewrite the integrand:

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

        2. Let u=1xu = \frac{1}{x}.

          Then let du=−dxx2du = - \frac{dx}{x^{2}} and substitute −du- du:

          ∫(−u2log⁡(1u)2−2ulog⁡(1u)+1u3) du\int \left(- \frac{u^{2} \log{\left(\frac{1}{u} \right)}^{2} - 2 u \log{\left(\frac{1}{u} \right)} + 1}{u^{3}}\right)\, du

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

            ∫u2log⁡(1u)2−2ulog⁡(1u)+1u3 du=−∫u2log⁡(1u)2−2ulog⁡(1u)+1u3 du\int \frac{u^{2} \log{\left(\frac{1}{u} \right)}^{2} - 2 u \log{\left(\frac{1}{u} \right)} + 1}{u^{3}}\, du = - \int \frac{u^{2} \log{\left(\frac{1}{u} \right)}^{2} - 2 u \log{\left(\frac{1}{u} \right)} + 1}{u^{3}}\, du

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

              Then let du=−duudu = - \frac{du}{u} and substitute dudu:

              ∫(−u2+2ueu−e2u) du\int \left(- u^{2} + 2 u e^{u} - e^{2 u}\right)\, du

              1. Integrate term-by-term:

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

                  ∫(−u2) du=−∫u2 du\int \left(- u^{2}\right)\, du = - \int u^{2}\, du

                  1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                    ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

                  So, the result is: −u33- \frac{u^{3}}{3}

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

                  ∫2ueu du=2∫ueu du\int 2 u e^{u}\, du = 2 \int u 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)=uu{\left(u \right)} = u and let dv⁡(u)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

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

                    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. The integral of the exponential function is itself.

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

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

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

                  ∫(−e2u) du=−∫e2u du\int \left(- e^{2 u}\right)\, du = - \int e^{2 u}\, du

                  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: −e2u2- \frac{e^{2 u}}{2}

                The result is: −u33+2ueu−e2u2−2eu- \frac{u^{3}}{3} + 2 u e^{u} - \frac{e^{2 u}}{2} - 2 e^{u}

              Now substitute uu back in:

              −log⁡(1u)33+2log⁡(1u)u−2u−12u2- \frac{\log{\left(\frac{1}{u} \right)}^{3}}{3} + \frac{2 \log{\left(\frac{1}{u} \right)}}{u} - \frac{2}{u} - \frac{1}{2 u^{2}}

            So, the result is: log⁡(1u)33−2log⁡(1u)u+2u+12u2\frac{\log{\left(\frac{1}{u} \right)}^{3}}{3} - \frac{2 \log{\left(\frac{1}{u} \right)}}{u} + \frac{2}{u} + \frac{1}{2 u^{2}}

          Now substitute uu back in:

          x22−2xlog⁡(x)+2x+log⁡(x)33\frac{x^{2}}{2} - 2 x \log{\left(x \right)} + 2 x + \frac{\log{\left(x \right)}^{3}}{3}

        Method #3

        1. Rewrite the integrand:

          (−x+log⁡(x))2x=x−2log⁡(x)+log⁡(x)2x\frac{\left(- x + \log{\left(x \right)}\right)^{2}}{x} = x - 2 \log{\left(x \right)} + \frac{\log{\left(x \right)}^{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:

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

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

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

            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)=1\operatorname{dv}{\left(x \right)} = 1.

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

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

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

                ∫1 dx=x\int 1\, dx = x

              Now evaluate the sub-integral.

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

              ∫1 dx=x\int 1\, dx = x

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

          1. Let u=1xu = \frac{1}{x}.

            Then let du=−dxx2du = - \frac{dx}{x^{2}} and substitute −du- du:

            ∫(−log⁡(1u)2u) du\int \left(- \frac{\log{\left(\frac{1}{u} \right)}^{2}}{u}\right)\, du

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

              ∫log⁡(1u)2u du=−∫log⁡(1u)2u du\int \frac{\log{\left(\frac{1}{u} \right)}^{2}}{u}\, du = - \int \frac{\log{\left(\frac{1}{u} \right)}^{2}}{u}\, du

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

                Then let du=−duudu = - \frac{du}{u} and substitute −du- du:

                ∫(−u2) du\int \left(- u^{2}\right)\, du

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

                  ∫u2 du=−∫u2 du\int u^{2}\, du = - \int u^{2}\, du

                  1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                    ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

                  So, the result is: −u33- \frac{u^{3}}{3}

                Now substitute uu back in:

                −log⁡(1u)33- \frac{\log{\left(\frac{1}{u} \right)}^{3}}{3}

              So, the result is: log⁡(1u)33\frac{\log{\left(\frac{1}{u} \right)}^{3}}{3}

            Now substitute uu back in:

            log⁡(x)33\frac{\log{\left(x \right)}^{3}}{3}

          The result is: x22−2xlog⁡(x)+2x+log⁡(x)33\frac{x^{2}}{2} - 2 x \log{\left(x \right)} + 2 x + \frac{\log{\left(x \right)}^{3}}{3}

      So, the result is: −x22+2xlog⁡(x)−2x−log⁡(x)33- \frac{x^{2}}{2} + 2 x \log{\left(x \right)} - 2 x - \frac{\log{\left(x \right)}^{3}}{3}

    The result is: 2xlog⁡(x)−2x−log⁡(x)332 x \log{\left(x \right)} - 2 x - \frac{\log{\left(x \right)}^{3}}{3}

  2. Add the constant of integration:

    2xlog⁡(x)−2x−log⁡(x)33+constant2 x \log{\left(x \right)} - 2 x - \frac{\log{\left(x \right)}^{3}}{3}+ \mathrm{constant}

The answer is:

2xlog⁡(x)−2x−log⁡(x)33+constant2 x \log{\left(x \right)} - 2 x - \frac{\log{\left(x \right)}^{3}}{3}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                       
 |                                                        
 | /                2\                   3                
 | |    (log(x) - x) |                log (x)             
 | |x - -------------| dx = C - 2*x - ------- + 2*x*log(x)
 | \          x      /                   3                
 |                                                        
/
∫(x−(−x+log⁡(x))2x) dx=C+2xlog⁡(x)−2x−log⁡(x)33\int \left(x - \frac{\left(- x + \log{\left(x \right)}\right)^{2}}{x}\right)\, dx = C + 2 x \log{\left(x \right)} - 2 x - \frac{\log{\left(x \right)}^{3}}{3}
Simplify
The answer [src] 
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−∞-\infty 
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−∞-\infty 
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Numerical answer [src] 
-28570.3797156332
-28570.3797156332

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

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