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Integral of d{x}:
Integral of sin
Integral of x/(x^4-16)
Integral of sin^3xcos^2xdx
Integral of 2sin4x
Graphing y =:
x/(x^4-16)
Identical expressions
x/(x^ four - sixteen)
x divide by (x to the power of 4 minus 16)
x divide by (x to the power of four minus sixteen)
x/(x4-16)
x/x4-16
x/(x⁴-16)
x/x^4-16
x divide by (x^4-16)
x/(x^4-16)dx
Similar expressions
x/(x^4+16)

Solving integrals/ x/(x^4-16)

Integral of x/(x^4-16) dx

The solution

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 oo           
  /           
 |            
 |     x      
 |  ------- dx
 |   4        
 |  x  - 16   
 |            
/             
6

$$\int\limits_{6}^{\infty} \frac{x}{x^{4} - 16}\, dx$$

Integral(x/(x^4 - 1*16), (x, 6, oo))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = x^{2}$ .
  
  Then let $du = 2 x dx$ and substitute $du$ :
  
  $\int \frac{1}{2 u^{2} - 32}\, du$
  1. Rewrite the integrand:
    
    $\frac{1}{2 u^{2} - 32} = \frac{- \frac{1}{u + 4} + \frac{1}{u - 4}}{16}$
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{- \frac{1}{u + 4} + \frac{1}{u - 4}}{16}\, du = \frac{\int \left(- \frac{1}{u + 4} + \frac{1}{u - 4}\right)\, du}{16}$
    1. Integrate term-by-term:
      
      The integral of $\frac{1}{u - 4}$ is $\log{\left(u - 4 \right)}$ .
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u + 4}\right)\, du = - \int \frac{1}{u + 4}\, du$
      
      The integral of $\frac{1}{u + 4}$ is $\log{\left(u + 4 \right)}$ .
      
      So, the result is: $- \log{\left(u + 4 \right)}$
      
      The result is: $\log{\left(u - 4 \right)} - \log{\left(u + 4 \right)}$
    So, the result is: $\frac{\log{\left(u - 4 \right)}}{16} - \frac{\log{\left(u + 4 \right)}}{16}$
  Now substitute $u$ back in:
  
  $\frac{\log{\left(x^{2} - 4 \right)}}{16} - \frac{\log{\left(x^{2} + 4 \right)}}{16}$
Method #2
1. Rewrite the integrand:
  
  $\frac{x}{x^{4} - 16} = - \frac{x}{8 \left(x^{2} + 4\right)} + \frac{1}{16 \left(x + 2\right)} + \frac{1}{16 \left(x - 2\right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{x}{8 \left(x^{2} + 4\right)}\right)\, dx = - \frac{\int \frac{x}{x^{2} + 4}\, dx}{8}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{x}{x^{2} + 4}\, dx = \frac{\int \frac{2 x}{x^{2} + 4}\, dx}{2}$
      
      Let $u = x^{2} + 4$ .
      
      Then let $du = 2 x dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{2 u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x^{2} + 4 \right)}$
      
      So, the result is: $\frac{\log{\left(x^{2} + 4 \right)}}{2}$
    So, the result is: $- \frac{\log{\left(x^{2} + 4 \right)}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{16 \left(x + 2\right)}\, dx = \frac{\int \frac{1}{x + 2}\, dx}{16}$
    1. Let $u = x + 2$ .
      
      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:
      
      $\log{\left(x + 2 \right)}$
    So, the result is: $\frac{\log{\left(x + 2 \right)}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{16 \left(x - 2\right)}\, dx = \frac{\int \frac{1}{x - 2}\, dx}{16}$
    1. Let $u = x - 2$ .
      
      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:
      
      $\log{\left(x - 2 \right)}$
    So, the result is: $\frac{\log{\left(x - 2 \right)}}{16}$
  The result is: $\frac{\log{\left(x - 2 \right)}}{16} + \frac{\log{\left(x + 2 \right)}}{16} - \frac{\log{\left(x^{2} + 4 \right)}}{16}$
Add the constant of integration:

$\frac{\log{\left(x^{2} - 4 \right)}}{16} - \frac{\log{\left(x^{2} + 4 \right)}}{16}+ \mathrm{constant}$

The answer is:

$\frac{\log{\left(x^{2} - 4 \right)}}{16} - \frac{\log{\left(x^{2} + 4 \right)}}{16}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                           
 |                     /     2\      /      2\
 |    x             log\4 + x /   log\-4 + x /
 | ------- dx = C - ----------- + ------------
 |  4                    16            16     
 | x  - 16                                    
 |                                            
/

$${{\log \left(x^2-4\right)}\over{16}}-{{\log \left(x^2+4\right) }\over{16}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  log(32)   log(40)
- ------- + -------
     16        16

$$- \frac{\log{\left(32 \right)}}{16} + \frac{\log{\left(40 \right)}}{16}$$

  log(32)   log(40)
- ------- + -------
     16        16

$$- \frac{\log{\left(32 \right)}}{16} + \frac{\log{\left(40 \right)}}{16}$$

Simplify

The graph

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

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Identical expressions

Similar expressions

Integral of x/(x^4-16) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2