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

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

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The solution

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

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

    Then let du=dxxdu = \frac{dx}{x} and substitute −du- du:

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

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

      ∫ue−u du=−∫ue−u du\int u e^{- u}\, du = - \int u e^{- u}\, du

      1. There are multiple ways to do this integral.

        Method #1

        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}

        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(u)=uu{\left(u \right)} = u and let dv⁡(u)=e−u\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. Let u=−uu = - u.

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

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

            Now substitute uu back in:

            −e−u- e^{- u}

          Now evaluate the sub-integral.

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

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

          1. Let u=−uu = - u.

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

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

            Now substitute uu back in:

            −e−u- e^{- u}

          So, the result is: e−ue^{- u}

      So, the result is: ue−u+e−uu e^{- u} + e^{- u}

    Now substitute uu back in:

    log⁡(x)x+1x\frac{\log{\left(x \right)}}{x} + \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] 
  /                            
 |                             
 | -log(x)           1   log(x)
 | -------- dx = C + - + ------
 |     2             x     x   
 |    x                        
 |                             
/
∫(−1)log⁡(x)x2 dx=C+log⁡(x)x+1x\int \frac{\left(-1\right) \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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