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

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

Limits of integration:

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The solution

You have entered [src] 
  1          
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∫01log⁡(x)x2 dx\int\limits_{0}^{1} \frac{\log{\left(x \right)}}{x^{2}}\, dx 
Integral(log(x)/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:

      ∫ue−u du\int u e^{- u}\, du

      1. Let u=−uu = - u.

        Then let du=−dudu = - du and substitute dudu:

        ∫ueu du\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}

        Now substitute uu back in:

        −ue−u−e−u- u e^{- u} - e^{- u}

      Now substitute uu back in:

      −log⁡(x)x−1x- \frac{\log{\left(x \right)}}{x} - \frac{1}{x}

    Method #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(x)=log⁡(x)u{\left(x \right)} = \log{\left(x \right)} and let dv⁡(x)=1x2\operatorname{dv}{\left(x \right)} = \frac{1}{x^{2}}.

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

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

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

        ∫1x2 dx=−1x\int \frac{1}{x^{2}}\, dx = - \frac{1}{x}

      Now evaluate the sub-integral.

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

      ∫(−1x2) dx=−∫1x2 dx\int \left(- \frac{1}{x^{2}}\right)\, dx = - \int \frac{1}{x^{2}}\, dx

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

        ∫1x2 dx=−1x\int \frac{1}{x^{2}}\, dx = - \frac{1}{x}

      So, the result is: 1x\frac{1}{x}

  2. Now simplify:

    −log⁡(x)+1x- \frac{\log{\left(x \right)} + 1}{x}

  3. Add the constant of integration:

    −log⁡(x)+1x+constant- \frac{\log{\left(x \right)} + 1}{x}+ \mathrm{constant}

The answer is:

−log⁡(x)+1x+constant- \frac{\log{\left(x \right)} + 1}{x}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                          
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 | log(x)          1   log(x)
 | ------ dx = C - - - ------
 |    2            x     x   
 |   x                       
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∫log⁡(x)x2 dx=C−log⁡(x)x−1x\int \frac{\log{\left(x \right)}}{x^{2}}\, dx = C - \frac{\log{\left(x \right)}}{x} - \frac{1}{x}
Simplify
The answer [src] 
-oo
−∞-\infty 
=
= 
-oo
−∞-\infty 
-oo
Numerical answer [src] 
-5.93814806236544e+20
-5.93814806236544e+20

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

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