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one /((×+ one)*(x+ one)*(x- one))
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1/((×+1)(x+1)(x-1))
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1/((×+1)*(x+1)*(x+1))
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1/((×+1)*(x-1)*(x-1))

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

Integral of 1/((×+1)(x+1)(x-1)) dx

The solution

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  1                           
  /                           
 |                            
 |             1              
 |  ----------------------- dx
 |  (x + 1)*(x + 1)*(x - 1)   
 |                            
/                             
0

$$\int\limits_{0}^{1} \frac{1}{\left(x + 1\right) \left(x + 1\right) \left(x - 1\right)}\, dx$$

Integral(1/(((x + 1)*(x + 1))*(x - 1)), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{1}{\left(x + 1\right) \left(x + 1\right) \left(x - 1\right)} = - \frac{1}{4 \left(x + 1\right)} - \frac{1}{2 \left(x + 1\right)^{2}} + \frac{1}{4 \left(x - 1\right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{4 \left(x + 1\right)}\right)\, dx = - \frac{\int \frac{1}{x + 1}\, dx}{4}$
    1. Let $u = x + 1$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x + 1 \right)}$
    So, the result is: $- \frac{\log{\left(x + 1 \right)}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{2 \left(x + 1\right)^{2}}\right)\, dx = - \frac{\int \frac{1}{\left(x + 1\right)^{2}}\, dx}{2}$
    1. Let $u = x + 1$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u^{2}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$
      
      Now substitute $u$ back in:
      
      $- \frac{1}{x + 1}$
    So, the result is: $\frac{1}{2 \left(x + 1\right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{4 \left(x - 1\right)}\, dx = \frac{\int \frac{1}{x - 1}\, dx}{4}$
    1. Let $u = x - 1$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x - 1 \right)}$
    So, the result is: $\frac{\log{\left(x - 1 \right)}}{4}$
  The result is: $\frac{\log{\left(x - 1 \right)}}{4} - \frac{\log{\left(x + 1 \right)}}{4} + \frac{1}{2 \left(x + 1\right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{1}{\left(x + 1\right) \left(x + 1\right) \left(x - 1\right)} = \frac{1}{x^{3} + x^{2} - x - 1}$
2. Rewrite the integrand:
  
  $\frac{1}{x^{3} + x^{2} - x - 1} = - \frac{1}{4 \left(x + 1\right)} - \frac{1}{2 \left(x + 1\right)^{2}} + \frac{1}{4 \left(x - 1\right)}$
3. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{4 \left(x + 1\right)}\right)\, dx = - \frac{\int \frac{1}{x + 1}\, dx}{4}$
    1. Let $u = x + 1$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x + 1 \right)}$
    So, the result is: $- \frac{\log{\left(x + 1 \right)}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{2 \left(x + 1\right)^{2}}\right)\, dx = - \frac{\int \frac{1}{\left(x + 1\right)^{2}}\, dx}{2}$
    1. Let $u = x + 1$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u^{2}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$
      
      Now substitute $u$ back in:
      
      $- \frac{1}{x + 1}$
    So, the result is: $\frac{1}{2 \left(x + 1\right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{4 \left(x - 1\right)}\, dx = \frac{\int \frac{1}{x - 1}\, dx}{4}$
    1. Let $u = x - 1$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x - 1 \right)}$
    So, the result is: $\frac{\log{\left(x - 1 \right)}}{4}$
  The result is: $\frac{\log{\left(x - 1 \right)}}{4} - \frac{\log{\left(x + 1 \right)}}{4} + \frac{1}{2 \left(x + 1\right)}$
Method #3
1. Rewrite the integrand:
  
  $\frac{1}{\left(x + 1\right) \left(x + 1\right) \left(x - 1\right)} = \frac{1}{x^{3} + x^{2} - x - 1}$
2. Rewrite the integrand:
  
  $\frac{1}{x^{3} + x^{2} - x - 1} = - \frac{1}{4 \left(x + 1\right)} - \frac{1}{2 \left(x + 1\right)^{2}} + \frac{1}{4 \left(x - 1\right)}$
3. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{4 \left(x + 1\right)}\right)\, dx = - \frac{\int \frac{1}{x + 1}\, dx}{4}$
    1. Let $u = x + 1$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x + 1 \right)}$
    So, the result is: $- \frac{\log{\left(x + 1 \right)}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{2 \left(x + 1\right)^{2}}\right)\, dx = - \frac{\int \frac{1}{\left(x + 1\right)^{2}}\, dx}{2}$
    1. Let $u = x + 1$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u^{2}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$
      
      Now substitute $u$ back in:
      
      $- \frac{1}{x + 1}$
    So, the result is: $\frac{1}{2 \left(x + 1\right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{4 \left(x - 1\right)}\, dx = \frac{\int \frac{1}{x - 1}\, dx}{4}$
    1. Let $u = x - 1$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x - 1 \right)}$
    So, the result is: $\frac{\log{\left(x - 1 \right)}}{4}$
  The result is: $\frac{\log{\left(x - 1 \right)}}{4} - \frac{\log{\left(x + 1 \right)}}{4} + \frac{1}{2 \left(x + 1\right)}$
Now simplify:

$\frac{\left(x + 1\right) \left(\log{\left(x - 1 \right)} - \log{\left(x + 1 \right)}\right) + 2}{4 \left(x + 1\right)}$
Add the constant of integration:

$\frac{\left(x + 1\right) \left(\log{\left(x - 1 \right)} - \log{\left(x + 1 \right)}\right) + 2}{4 \left(x + 1\right)}+ \mathrm{constant}$

The answer is:

$\frac{\left(x + 1\right) \left(\log{\left(x - 1 \right)} - \log{\left(x + 1 \right)}\right) + 2}{4 \left(x + 1\right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                     
 |                                                                      
 |            1                         1       log(1 + x)   log(-1 + x)
 | ----------------------- dx = C + --------- - ---------- + -----------
 | (x + 1)*(x + 1)*(x - 1)          2*(1 + x)       4             4     
 |                                                                      
/

$$\int \frac{1}{\left(x + 1\right) \left(x + 1\right) \left(x - 1\right)}\, dx = C + \frac{\log{\left(x - 1 \right)}}{4} - \frac{\log{\left(x + 1 \right)}}{4} + \frac{1}{2 \left(x + 1\right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

      pi*I
-oo - ----
       4

$$-\infty - \frac{i \pi}{4}$$

      pi*I
-oo - ----
       4

$$-\infty - \frac{i \pi}{4}$$

-oo - pi*i/4

Numerical answer [src]

-11.4460259916949

-11.4460259916949

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of 1/((×+1)(x+1)(x-1)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of 1/((×+1)*(x+1)*(x-1)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Integral of 1/((×+1)(x+1)(x-1)) dx