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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 \left(- \frac{2}{x + 1}\right)\, 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. The integral of a constant times a function is the constant times the integral of the function:

          ∫(−1x+1) dx=−∫1x+1 dx\int \left(- \frac{1}{x + 1}\right)\, dx = - \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: −log⁡(x+1)- \log{\left(x + 1 \right)}

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

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

        ∫(−1x+1) dx=−∫1x+1 dx\int \left(- \frac{1}{x + 1}\right)\, dx = - \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: −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] 
  /                               
 |                                
 | 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
0.001.000.100.200.300.400.500.600.700.800.901-2
Plot the graph f(x)
The answer [src] 
1 - 2*log(2)
1−2log⁡(2)1 - 2 \log{\left(2 \right)} 
=
= 
1 - 2*log(2)
1−2log⁡(2)1 - 2 \log{\left(2 \right)} 
1 - 2*log(2)
Numerical answer [src] 
-0.386294361119891
-0.386294361119891
The graph
Integral of (x-1)/(x+1) dx

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

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