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x/x√1-ln^2x

Integral of x/x√1-ln^2x dx

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The solution

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12(log(x)2+1xx)dx\int\limits_{1}^{\sqrt{2}} \left(- \log{\left(x \right)}^{2} + \frac{\sqrt{1} x}{x}\right)\, dx
Integral(x*sqrt(1)/x - log(x)^2, (x, 1, sqrt(2)))
Detail solution
  1. Integrate term-by-term:

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

      (log(x)2)dx=log(x)2dx\int \left(- \log{\left(x \right)}^{2}\right)\, dx = - \int \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:

        u2eudu\int u^{2} e^{u}\, du

        1. Use integration by parts:

          udv=uvvdu\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.

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

          Now evaluate the sub-integral.

        2. Use integration by parts:

          udv=uvvdu\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.

            eudu=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:

          2eudu=2eudu\int 2 e^{u}\, du = 2 \int e^{u}\, du

          1. The integral of the exponential function is itself.

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

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

        Now substitute uu back in:

        xlog(x)22xlog(x)+2xx \log{\left(x \right)}^{2} - 2 x \log{\left(x \right)} + 2 x

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

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

      1xxdx=xxdx\int \frac{\sqrt{1} x}{x}\, dx = \int \frac{x}{x}\, dx

      1. Don't know the steps in finding this integral.

        But the integral is

        xx

      So, the result is: xx

    The result is: xlog(x)2+2xlog(x)x- x \log{\left(x \right)}^{2} + 2 x \log{\left(x \right)} - x

  2. Now simplify:

    x(log(x)2+2log(x)1)x \left(- \log{\left(x \right)}^{2} + 2 \log{\left(x \right)} - 1\right)

  3. Add the constant of integration:

    x(log(x)2+2log(x)1)+constantx \left(- \log{\left(x \right)}^{2} + 2 \log{\left(x \right)} - 1\right)+ \mathrm{constant}


The answer is:

x(log(x)2+2log(x)1)+constantx \left(- \log{\left(x \right)}^{2} + 2 \log{\left(x \right)} - 1\right)+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                                       
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 | |x*\/ 1       2   |                   2                
 | |------- - log (x)| dx = C - x - x*log (x) + 2*x*log(x)
 | \   x             /                                    
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xx((logx)22logx+2)x-x\,\left(\left(\log x\right)^2-2\,\log x+2\right)
The graph
1.001.051.101.151.201.251.301.351.402-2
The answer [src]
      ___     ___    2/  ___\       ___    /  ___\
1 - \/ 2  - \/ 2 *log \\/ 2 / + 2*\/ 2 *log\\/ 2 /
(log2)24log2232+4232-{{\left(\log 2\right)^2-4\,\log 2-2^{{{3}\over{2}}}+4}\over{2^{{{3 }\over{2}}}}}
=
=
      ___     ___    2/  ___\       ___    /  ___\
1 - \/ 2  - \/ 2 *log \\/ 2 / + 2*\/ 2 *log\\/ 2 /
22log(2)2+22log(2)+1- \sqrt{2} - \sqrt{2} \log{\left(\sqrt{2} \right)}^{2} + 2 \sqrt{2} \log{\left(\sqrt{2} \right)} + 1
Numerical answer [src]
0.396178789003915
0.396178789003915
The graph
Integral of x/x√1-ln^2x dx

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