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xdx/(x^2-1)^3

Solving integrals/ xdx/(x^2+1)^3

Integral of xdx/(x^2+1)^3 dx

The solution

You have entered [src]

  1             
  /             
 |              
 |      x       
 |  --------- dx
 |          3   
 |  / 2    \    
 |  \x  + 1/    
 |              
/               
0

$$\int\limits_{0}^{1} \frac{x}{\left(x^{2} + 1\right)^{3}}\, dx$$

Integral(x/(x^2 + 1)^3, (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{x}{\left(x^{2} + 1\right)^{3}} = \frac{x}{x^{6} + 3 x^{4} + 3 x^{2} + 1}$
2. Let $u = x^{2}$ .
  
  Then let $du = 2 x dx$ and substitute $du$ :
  
  $\int \frac{1}{2 u^{3} + 6 u^{2} + 6 u + 2}\, du$
  1. Rewrite the integrand:
    
    $\frac{1}{2 u^{3} + 6 u^{2} + 6 u + 2} = \frac{1}{2 \left(u + 1\right)^{3}}$
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{2 \left(u + 1\right)^{3}}\, du = \frac{\int \frac{1}{\left(u + 1\right)^{3}}\, du}{2}$
    1. Let $u = u + 1$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{u^{3}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{3}}\, du = - \frac{1}{2 u^{2}}$
      
      Now substitute $u$ back in:
      
      $- \frac{1}{2 \left(u + 1\right)^{2}}$
    So, the result is: $- \frac{1}{4 \left(u + 1\right)^{2}}$
  Now substitute $u$ back in:
  
  $- \frac{1}{4 \left(x^{2} + 1\right)^{2}}$
Method #2
1. Rewrite the integrand:
  
  $\frac{x}{\left(x^{2} + 1\right)^{3}} = \frac{x}{x^{6} + 3 x^{4} + 3 x^{2} + 1}$
2. Let $u = x^{2}$ .
  
  Then let $du = 2 x dx$ and substitute $du$ :
  
  $\int \frac{1}{2 u^{3} + 6 u^{2} + 6 u + 2}\, du$
  1. Rewrite the integrand:
    
    $\frac{1}{2 u^{3} + 6 u^{2} + 6 u + 2} = \frac{1}{2 \left(u + 1\right)^{3}}$
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{2 \left(u + 1\right)^{3}}\, du = \frac{\int \frac{1}{\left(u + 1\right)^{3}}\, du}{2}$
    1. Let $u = u + 1$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{u^{3}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{3}}\, du = - \frac{1}{2 u^{2}}$
      
      Now substitute $u$ back in:
      
      $- \frac{1}{2 \left(u + 1\right)^{2}}$
    So, the result is: $- \frac{1}{4 \left(u + 1\right)^{2}}$
  Now substitute $u$ back in:
  
  $- \frac{1}{4 \left(x^{2} + 1\right)^{2}}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

  /                              
 |                               
 |     x                   1     
 | --------- dx = C - -----------
 |         3                    2
 | / 2    \             /     2\ 
 | \x  + 1/           4*\1 + x / 
 |                               
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

3/16

$$\frac{3}{16}$$

3/16

$$\frac{3}{16}$$

3/16

Numerical answer [src]

0.1875

0.1875

The graph

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

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Identical expressions

Similar expressions

Integral of xdx/(x^2+1)^3 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2