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Integral of (1+ln(x-1))/(x-1) dx

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  1                  
  /                  
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 |  1 + log(x - 1)   
 |  -------------- dx
 |      x - 1        
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-1                   
11log(x1)+1x1dx\int\limits_{-1}^{1} \frac{\log{\left(x - 1 \right)} + 1}{x - 1}\, dx
Integral((1 + log(x - 1))/(x - 1), (x, -1, 1))
Detail solution
  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=log(x1)+1u = \log{\left(x - 1 \right)} + 1.

      Then let du=dxx1du = \frac{dx}{x - 1} and substitute dudu:

      udu\int u\, du

      1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

        udu=u22\int u\, du = \frac{u^{2}}{2}

      Now substitute uu back in:

      (log(x1)+1)22\frac{\left(\log{\left(x - 1 \right)} + 1\right)^{2}}{2}

    Method #2

    1. Rewrite the integrand:

      log(x1)+1x1=log(x1)x1+1x1\frac{\log{\left(x - 1 \right)} + 1}{x - 1} = \frac{\log{\left(x - 1 \right)}}{x - 1} + \frac{1}{x - 1}

    2. Integrate term-by-term:

      1. Let u=x1u = x - 1.

        Then let du=dxdu = dx and substitute dudu:

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

        1. Let u=1uu = \frac{1}{u}.

          Then let du=duu2du = - \frac{du}{u^{2}} and substitute du- du:

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

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

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

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

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

              (u)du\int \left(- u\right)\, du

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

                udu=udu\int u\, du = - \int u\, du

                1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

                  udu=u22\int u\, du = \frac{u^{2}}{2}

                So, the result is: u22- \frac{u^{2}}{2}

              Now substitute uu back in:

              log(1u)22- \frac{\log{\left(\frac{1}{u} \right)}^{2}}{2}

            So, the result is: log(1u)22\frac{\log{\left(\frac{1}{u} \right)}^{2}}{2}

          Now substitute uu back in:

          log(u)22\frac{\log{\left(u \right)}^{2}}{2}

        Now substitute uu back in:

        log(x1)22\frac{\log{\left(x - 1 \right)}^{2}}{2}

      1. Let u=x1u = x - 1.

        Then let du=dxdu = dx and substitute dudu:

        1udu\int \frac{1}{u}\, du

        1. The integral of 1u\frac{1}{u} is log(u)\log{\left(u \right)}.

        Now substitute uu back in:

        log(x1)\log{\left(x - 1 \right)}

      The result is: log(x1)22+log(x1)\frac{\log{\left(x - 1 \right)}^{2}}{2} + \log{\left(x - 1 \right)}

  2. Now simplify:

    (log(x1)+1)22\frac{\left(\log{\left(x - 1 \right)} + 1\right)^{2}}{2}

  3. Add the constant of integration:

    (log(x1)+1)22+constant\frac{\left(\log{\left(x - 1 \right)} + 1\right)^{2}}{2}+ \mathrm{constant}


The answer is:

(log(x1)+1)22+constant\frac{\left(\log{\left(x - 1 \right)} + 1\right)^{2}}{2}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                         
 |                                         2
 | 1 + log(x - 1)          (1 + log(x - 1)) 
 | -------------- dx = C + -----------------
 |     x - 1                       2        
 |                                          
/                                           
log(x1)+1x1dx=C+(log(x1)+1)22\int \frac{\log{\left(x - 1 \right)} + 1}{x - 1}\, dx = C + \frac{\left(\log{\left(x - 1 \right)} + 1\right)^{2}}{2}
The answer [src]
                    2       
     (pi*I + log(2))        
oo - ---------------- - pi*I
            2               
iπ(log(2)+iπ)22\infty - i \pi - \frac{\left(\log{\left(2 \right)} + i \pi\right)^{2}}{2}
=
=
                    2       
     (pi*I + log(2))        
oo - ---------------- - pi*I
            2               
iπ(log(2)+iπ)22\infty - i \pi - \frac{\left(\log{\left(2 \right)} + i \pi\right)^{2}}{2}
oo - (pi*i + log(2))^2/2 - pi*i
Numerical answer [src]
(897.312398743146 - 138.514234126527j)
(897.312398743146 - 138.514234126527j)

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