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

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

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The solution

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∫01log⁡(x)−1x2 dx\int\limits_{0}^{1} \frac{\log{\left(x \right)} - 1}{x^{2}}\, dx 
Integral((log(x) - 1)/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:

      ∫(u−1)e−u du\int \left(u - 1\right) e^{- u}\, du

      1. Let u=−uu = - u.

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

        ∫(ueu+eu) du\int \left(u e^{u} + e^{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)=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}

          1. The integral of the exponential function is itself.

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

          The result is: ueuu e^{u}

        Now substitute uu back in:

        −ue−u- u e^{- u}

      Now substitute uu back in:

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

    Method #2

    1. Rewrite the integrand:

      log⁡(x)−1x2=log⁡(x)x2−1x2\frac{\log{\left(x \right)} - 1}{x^{2}} = \frac{\log{\left(x \right)}}{x^{2}} - \frac{1}{x^{2}}

    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:

        ∫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}

      1. 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}

      The result is: −log⁡(x)x- \frac{\log{\left(x \right)}}{x}

  2. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                          
 |                           
 | log(x) - 1          log(x)
 | ---------- dx = C - ------
 |      2                x   
 |     x                     
 |                           
/
∫log⁡(x)−1x2 dx=C−log⁡(x)x\int \frac{\log{\left(x \right)} - 1}{x^{2}}\, dx = C - \frac{\log{\left(x \right)}}{x}
Simplify
The answer [src] 
-oo
−∞-\infty 
=
= 
-oo
−∞-\infty 
-oo
Numerical answer [src] 
-6.0760804301603e+20
-6.0760804301603e+20

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

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