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Limit of the function:
x^4/(1-x^2)
Identical expressions
x^ four /(one -x^ two)
x to the power of 4 divide by (1 minus x squared )
x to the power of four divide by (one minus x to the power of two)
x4/(1-x2)
x4/1-x2
x⁴/(1-x²)
x to the power of 4/(1-x to the power of 2)
x^4/1-x^2
x^4 divide by (1-x^2)
x^4/(1-x^2)dx
Similar expressions
x^4/(1+x^2)

Solving integrals/ x^4/(1-x^2)

Integral of x^4/(1-x^2) dx

The solution

You have entered [src]

  1          
  /          
 |           
 |     4     
 |    x      
 |  ------ dx
 |       2   
 |  1 - x    
 |           
/            
0

$$\int\limits_{0}^{1} \frac{x^{4}}{1 - x^{2}}\, dx$$

Integral(x^4/(1 - x^2), (x, 0, 1))

Detail solution

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

$- \frac{x^{3}}{3} - x - \frac{\log{\left(x - 1 \right)}}{2} + \frac{\log{\left(x + 1 \right)}}{2}+ \mathrm{constant}$

The answer is:

$- \frac{x^{3}}{3} - x - \frac{\log{\left(x - 1 \right)}}{2} + \frac{\log{\left(x + 1 \right)}}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                 
 |                                                  
 |    4                                            3
 |   x             log(1 + x)       log(-1 + x)   x 
 | ------ dx = C + ---------- - x - ----------- - --
 |      2              2                 2        3 
 | 1 - x                                            
 |                                                  
/

$$\int \frac{x^{4}}{1 - x^{2}}\, dx = C - \frac{x^{3}}{3} - x - \frac{\log{\left(x - 1 \right)}}{2} + \frac{\log{\left(x + 1 \right)}}{2}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

     pi*I
oo + ----
      2

$$\infty + \frac{i \pi}{2}$$

     pi*I
oo + ----
      2

$$\infty + \frac{i \pi}{2}$$

oo + pi*i/2

Numerical answer [src]

21.0587186500535

21.0587186500535

The graph

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of x^4/(1-x^2) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2