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Solving integrals/ ln(n)/n^2

Integral of ln(n)/n^2 dx

Limits of integration:

from to
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The graph:

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Piecewise:

The solution

You have entered [src] 
 oo          
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 |  log(n)   
 |  ------ dn
 |     2     
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∫1∞log⁡(n)n2 dn\int\limits_{1}^{\infty} \frac{\log{\left(n \right)}}{n^{2}}\, dn 
Integral(log(n)/n^2, (n, 1, oo))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

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

      Then let du=dnndu = \frac{dn}{n} 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⁡(n)n−1n- \frac{\log{\left(n \right)}}{n} - \frac{1}{n}

    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(n)=log⁡(n)u{\left(n \right)} = \log{\left(n \right)} and let dv⁡(n)=1n2\operatorname{dv}{\left(n \right)} = \frac{1}{n^{2}}.

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

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

      1. The integral of nnn^{n} is nn+1n+1\frac{n^{n + 1}}{n + 1} when n≠−1n \neq -1:

        ∫1n2 dn=−1n\int \frac{1}{n^{2}}\, dn = - \frac{1}{n}

      Now evaluate the sub-integral.

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

      ∫(−1n2) dn=−∫1n2 dn\int \left(- \frac{1}{n^{2}}\right)\, dn = - \int \frac{1}{n^{2}}\, dn

      1. The integral of nnn^{n} is nn+1n+1\frac{n^{n + 1}}{n + 1} when n≠−1n \neq -1:

        ∫1n2 dn=−1n\int \frac{1}{n^{2}}\, dn = - \frac{1}{n}

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

  2. Now simplify:

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

  3. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                          
 |                           
 | log(n)          1   log(n)
 | ------ dn = C - - - ------
 |    2            n     n   
 |   n                       
 |                           
/
∫log⁡(n)n2 dn=C−log⁡(n)n−1n\int \frac{\log{\left(n \right)}}{n^{2}}\, dn = C - \frac{\log{\left(n \right)}}{n} - \frac{1}{n}
Simplify
The graph
1.00001.01001.00101.00201.00301.00401.00501.00601.00701.00801.00901-2
Plot the graph f(x)
The answer [src] 
1
11 
=
= 
1
11 
1

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

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