Mister Exam

Integral of ln(ln(x)) dx

Limits of integration:

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The graph:

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Piecewise:

The solution

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01log(log(x))dx\int\limits_{0}^{1} \log{\left(\log{\left(x \right)} \right)}\, dx
Integral(log(log(x)), (x, 0, 1))
Detail solution
  1. Let u=log(x)u = \log{\left(x \right)}.

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

    eulog(u)du\int e^{u} \log{\left(u \right)}\, du

    1. There are multiple ways to do this integral.

      Method #1

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

        Then let du=duudu = \frac{du}{u} and substitute dudu:

        ueueeudu\int u e^{u} e^{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)=eueeu\operatorname{dv}{\left(u \right)} = e^{u} e^{e^{u}}.

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

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

          1. Let u=euu = e^{u}.

            Then let du=eududu = e^{u} du and substitute dudu:

            eudu\int e^{u}\, du

            1. The integral of the exponential function is itself.

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

            Now substitute uu back in:

            eeue^{e^{u}}

          Now evaluate the sub-integral.

        2. Let u=euu = e^{u}.

          Then let du=eududu = e^{u} du and substitute dudu:

          euudu\int \frac{e^{u}}{u}\, du

            EiRule(a=1, b=0, context=exp(_u)/_u, symbol=_u)

          Now substitute uu back in:

          Ei(eu)\operatorname{Ei}{\left(e^{u} \right)}

        Now substitute uu back in:

        eulog(u)Ei(u)e^{u} \log{\left(u \right)} - \operatorname{Ei}{\left(u \right)}

      Method #2

      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)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

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

        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.

        EiRule(a=1, b=0, context=exp(_u)/_u, symbol=_u)

    Now substitute uu back in:

    xlog(log(x))Ei(log(x))x \log{\left(\log{\left(x \right)} \right)} - \operatorname{Ei}{\left(\log{\left(x \right)} \right)}

  2. Add the constant of integration:

    xlog(log(x))Ei(log(x))+constantx \log{\left(\log{\left(x \right)} \right)} - \operatorname{Ei}{\left(\log{\left(x \right)} \right)}+ \mathrm{constant}


The answer is:

xlog(log(x))Ei(log(x))+constantx \log{\left(\log{\left(x \right)} \right)} - \operatorname{Ei}{\left(\log{\left(x \right)} \right)}+ \mathrm{constant}

The answer (Indefinite) [src]
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 | log(log(x)) dx = C - Ei(log(x)) + x*log(log(x))
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log(log(x))dx=C+xlog(log(x))Ei(log(x))\int \log{\left(\log{\left(x \right)} \right)}\, dx = C + x \log{\left(\log{\left(x \right)} \right)} - \operatorname{Ei}{\left(\log{\left(x \right)} \right)}
The graph
0.001.000.100.200.300.400.500.600.700.800.900.02-0.02
The answer [src]
-EulerGamma
γ- \gamma
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-EulerGamma
γ- \gamma
-EulerGamma
Numerical answer [src]
(-0.577215664901533 + 3.14159265358979j)
(-0.577215664901533 + 3.14159265358979j)
The graph
Integral of ln(ln(x)) dx

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