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1/(1-x^2)^2
Graphing y =:
1/(1-x^2)^2
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Solving integrals/ 1/(1-x^2)^2

Integral of 1/(1-x^2)^2 dx

The solution

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  1             
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 |  --------- dx
 |          2   
 |  /     2\    
 |  \1 - x /    
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0

$$\int\limits_{0}^{1} \frac{1}{\left(1 - x^{2}\right)^{2}}\, dx$$

Integral(1/((1 - x^2)^2), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{1}{\left(1 - x^{2}\right)^{2}} = \frac{1}{4 \left(x + 1\right)} + \frac{1}{4 \left(x + 1\right)^{2}} - \frac{1}{4 \left(x - 1\right)} + \frac{1}{4 \left(x - 1\right)^{2}}$
2. Integrate term-by-term:
  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: $\frac{\log{\left(x + 1 \right)}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{4 \left(x + 1\right)^{2}}\, dx = \frac{\int \frac{1}{\left(x + 1\right)^{2}}\, dx}{4}$
    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: $- \frac{1}{4 \left(x + 1\right)}$
  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:
    
    $\int \frac{1}{4 \left(x - 1\right)^{2}}\, dx = \frac{\int \frac{1}{\left(x - 1\right)^{2}}\, dx}{4}$
    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: $- \frac{1}{4 \left(x - 1\right)}$
  The result is: $- \frac{\log{\left(x - 1 \right)}}{4} + \frac{\log{\left(x + 1 \right)}}{4} - \frac{1}{4 \left(x + 1\right)} - \frac{1}{4 \left(x - 1\right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{1}{\left(1 - x^{2}\right)^{2}} = \frac{1}{x^{4} - 2 x^{2} + 1}$
2. Rewrite the integrand:
  
  $\frac{1}{x^{4} - 2 x^{2} + 1} = \frac{1}{4 \left(x + 1\right)} + \frac{1}{4 \left(x + 1\right)^{2}} - \frac{1}{4 \left(x - 1\right)} + \frac{1}{4 \left(x - 1\right)^{2}}$
3. Integrate term-by-term:
  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: $\frac{\log{\left(x + 1 \right)}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{4 \left(x + 1\right)^{2}}\, dx = \frac{\int \frac{1}{\left(x + 1\right)^{2}}\, dx}{4}$
    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: $- \frac{1}{4 \left(x + 1\right)}$
  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:
    
    $\int \frac{1}{4 \left(x - 1\right)^{2}}\, dx = \frac{\int \frac{1}{\left(x - 1\right)^{2}}\, dx}{4}$
    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: $- \frac{1}{4 \left(x - 1\right)}$
  The result is: $- \frac{\log{\left(x - 1 \right)}}{4} + \frac{\log{\left(x + 1 \right)}}{4} - \frac{1}{4 \left(x + 1\right)} - \frac{1}{4 \left(x - 1\right)}$
Method #3
1. Rewrite the integrand:
  
  $\frac{1}{\left(1 - x^{2}\right)^{2}} = \frac{1}{x^{4} - 2 x^{2} + 1}$
2. Rewrite the integrand:
  
  $\frac{1}{x^{4} - 2 x^{2} + 1} = \frac{1}{4 \left(x + 1\right)} + \frac{1}{4 \left(x + 1\right)^{2}} - \frac{1}{4 \left(x - 1\right)} + \frac{1}{4 \left(x - 1\right)^{2}}$
3. Integrate term-by-term:
  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: $\frac{\log{\left(x + 1 \right)}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{4 \left(x + 1\right)^{2}}\, dx = \frac{\int \frac{1}{\left(x + 1\right)^{2}}\, dx}{4}$
    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: $- \frac{1}{4 \left(x + 1\right)}$
  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:
    
    $\int \frac{1}{4 \left(x - 1\right)^{2}}\, dx = \frac{\int \frac{1}{\left(x - 1\right)^{2}}\, dx}{4}$
    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: $- \frac{1}{4 \left(x - 1\right)}$
  The result is: $- \frac{\log{\left(x - 1 \right)}}{4} + \frac{\log{\left(x + 1 \right)}}{4} - \frac{1}{4 \left(x + 1\right)} - \frac{1}{4 \left(x - 1\right)}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

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]

3.45048902814162e+18

3.45048902814162e+18

The graph

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

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Identical expressions

Similar expressions

Integral of 1/(1-x^2)^2 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3