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ln(1/x)
Solving integrals/ ln(1/x)

Integral of ln(1/x) dx

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

  1. There are multiple ways to do this integral.

    Method #1

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

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

      ∫(−log⁡(u)u2) du\int \left(- \frac{\log{\left(u \right)}}{u^{2}}\right)\, du

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

        ∫log⁡(u)u2 du=−∫log⁡(u)u2 du\int \frac{\log{\left(u \right)}}{u^{2}}\, du = - \int \frac{\log{\left(u \right)}}{u^{2}}\, 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:

            ∫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⁡(u)u−1u- \frac{\log{\left(u \right)}}{u} - \frac{1}{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)=log⁡(u)u{\left(u \right)} = \log{\left(u \right)} and let dv⁡(u)=1u2\operatorname{dv}{\left(u \right)} = \frac{1}{u^{2}}.

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

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

            1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

              ∫1u2 du=−1u\int \frac{1}{u^{2}}\, du = - \frac{1}{u}

            Now evaluate the sub-integral.

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

            ∫(−1u2) du=−∫1u2 du\int \left(- \frac{1}{u^{2}}\right)\, du = - \int \frac{1}{u^{2}}\, du

            1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

              ∫1u2 du=−1u\int \frac{1}{u^{2}}\, du = - \frac{1}{u}

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

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

      Now substitute uu back in:

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

    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(x)=log⁡(1x)u{\left(x \right)} = \log{\left(\frac{1}{x} \right)} and let dv⁡(x)=1\operatorname{dv}{\left(x \right)} = 1.

      Then du⁡(x)=−1x\operatorname{du}{\left(x \right)} = - \frac{1}{x}.

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

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

        ∫1 dx=x\int 1\, dx = x

      Now evaluate the sub-integral.

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

      ∫(−1) dx=−x\int \left(-1\right)\, dx = - x

  2. Now simplify:

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

  3. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
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∫log⁡(1x) dx=C−xlog⁡(x)+x\int \log{\left(\frac{1}{x} \right)}\, dx = C - x \log{\left(x \right)} + x
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The answer [src] 
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Numerical answer [src] 
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Integral of ln(1/x) dx

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

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