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(x+1)/(x-1)
Solving integrals/ (x+1)/(x-1)

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

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The solution

You have entered [src] 
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∫01x+1x−1 dx\int\limits_{0}^{1} \frac{x + 1}{x - 1}\, dx 
Integral((x + 1)/(x - 1), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      x+1x−1=1+2x−1\frac{x + 1}{x - 1} = 1 + \frac{2}{x - 1}

    2. Integrate term-by-term:

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

        ∫1 dx=x\int 1\, dx = x

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

        ∫2x−1 dx=2∫1x−1 dx\int \frac{2}{x - 1}\, dx = 2 \int \frac{1}{x - 1}\, dx

        1. Let u=x−1u = x - 1.

          Then let du=dxdu = dx and substitute dudu:

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

        So, the result is: 2log⁡(x−1)2 \log{\left(x - 1 \right)}

      The result is: x+2log⁡(x−1)x + 2 \log{\left(x - 1 \right)}

    Method #2

    1. Rewrite the integrand:

      x+1x−1=xx−1+1x−1\frac{x + 1}{x - 1} = \frac{x}{x - 1} + \frac{1}{x - 1}

    2. Integrate term-by-term:

      1. Rewrite the integrand:

        xx−1=1+1x−1\frac{x}{x - 1} = 1 + \frac{1}{x - 1}

      2. Integrate term-by-term:

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

          ∫1 dx=x\int 1\, dx = x

        1. Let u=x−1u = x - 1.

          Then let du=dxdu = dx and substitute dudu:

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

        The result is: x+log⁡(x−1)x + \log{\left(x - 1 \right)}

      1. Let u=x−1u = x - 1.

        Then let du=dxdu = dx and substitute dudu:

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

      The result is: x+log⁡(x−1)+log⁡(x−1)x + \log{\left(x - 1 \right)} + \log{\left(x - 1 \right)}

  2. Add the constant of integration:

    x+2log⁡(x−1)+constantx + 2 \log{\left(x - 1 \right)}+ \mathrm{constant}

The answer is:

x+2log⁡(x−1)+constantx + 2 \log{\left(x - 1 \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                
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 | x + 1                           
 | ----- dx = C + x + 2*log(-1 + x)
 | x - 1                           
 |                                 
/
∫x+1x−1 dx=C+x+2log⁡(x−1)\int \frac{x + 1}{x - 1}\, dx = C + x + 2 \log{\left(x - 1 \right)}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90-2500025000
Plot the graph f(x)
The answer [src] 
-oo - 2*pi*I
−∞−2iπ-\infty - 2 i \pi 
=
= 
-oo - 2*pi*I
−∞−2iπ-\infty - 2 i \pi 
-oo - 2*pi*i
Numerical answer [src] 
-87.181913572439
-87.181913572439
The graph
Integral of (x+1)/(x-1) dx

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

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