Mister Exam

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

Limits of integration:

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Piecewise:

The solution

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  1               
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 |  log(log(x))   
 |  ----------- dx
 |       x        
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01log(log(x))xdx\int\limits_{0}^{1} \frac{\log{\left(\log{\left(x \right)} \right)}}{x}\, dx
Integral(log(log(x))/x, (x, 0, 1))
Detail solution
  1. There are multiple ways to do this integral.

    Method #1

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

      Then let du=dxxlog(x)du = \frac{dx}{x \log{\left(x \right)}} and substitute dudu:

      ueudu\int u 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)=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.

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

        Now evaluate the sub-integral.

      2. The integral of the exponential function is itself.

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

      Now substitute uu back in:

      log(x)log(log(x))log(x)\log{\left(x \right)} \log{\left(\log{\left(x \right)} \right)} - \log{\left(x \right)}

    Method #2

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

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

      log(u)du\int \log{\left(u \right)}\, 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)=log(u)u{\left(u \right)} = \log{\left(u \right)} and let dv(u)=1\operatorname{dv}{\left(u \right)} = 1.

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

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

        1. The integral of a constant is the constant times the variable of integration:

          1du=u\int 1\, du = u

        Now evaluate the sub-integral.

      2. The integral of a constant is the constant times the variable of integration:

        1du=u\int 1\, du = u

      Now substitute uu back in:

      log(x)log(log(x))log(x)\log{\left(x \right)} \log{\left(\log{\left(x \right)} \right)} - \log{\left(x \right)}

    Method #3

    1. Let u=1xu = \frac{1}{x}.

      Then let du=dxx2du = - \frac{dx}{x^{2}} and substitute du- du:

      (log(log(1u))u)du\int \left(- \frac{\log{\left(\log{\left(\frac{1}{u} \right)} \right)}}{u}\right)\, du

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

        log(log(1u))udu=log(log(1u))udu\int \frac{\log{\left(\log{\left(\frac{1}{u} \right)} \right)}}{u}\, du = - \int \frac{\log{\left(\log{\left(\frac{1}{u} \right)} \right)}}{u}\, du

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

          Then let du=duulog(1u)du = - \frac{du}{u \log{\left(\frac{1}{u} \right)}} and substitute du- du:

          (ueu)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:

            ueudu=ueudu\int u e^{u}\, du = - \int u 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)=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.

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

              Now evaluate the sub-integral.

            2. The integral of the exponential function is itself.

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

            So, the result is: ueu+eu- u e^{u} + e^{u}

          Now substitute uu back in:

          log(1u)log(log(1u))+log(1u)- \log{\left(\frac{1}{u} \right)} \log{\left(\log{\left(\frac{1}{u} \right)} \right)} + \log{\left(\frac{1}{u} \right)}

        So, the result is: log(1u)log(log(1u))log(1u)\log{\left(\frac{1}{u} \right)} \log{\left(\log{\left(\frac{1}{u} \right)} \right)} - \log{\left(\frac{1}{u} \right)}

      Now substitute uu back in:

      log(x)log(log(x))log(x)\log{\left(x \right)} \log{\left(\log{\left(x \right)} \right)} - \log{\left(x \right)}

  2. Now simplify:

    (log(log(x))1)log(x)\left(\log{\left(\log{\left(x \right)} \right)} - 1\right) \log{\left(x \right)}

  3. Add the constant of integration:

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


The answer is:

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

The answer (Indefinite) [src]
  /                                                
 |                                                 
 | log(log(x))                                     
 | ----------- dx = C - log(x) + log(x)*log(log(x))
 |      x                                          
 |                                                 
/                                                  
log(log(x))xdx=C+log(x)log(log(x))log(x)\int \frac{\log{\left(\log{\left(x \right)} \right)}}{x}\, dx = C + \log{\left(x \right)} \log{\left(\log{\left(x \right)} \right)} - \log{\left(x \right)}
The answer [src]
oo
\infty
=
=
oo
\infty
oo
Numerical answer [src]
(122.846251720628 + 138.514221668049j)
(122.846251720628 + 138.514221668049j)

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