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1/(x^4-16)
Identical expressions
one /(x^ four - sixteen)
1 divide by (x to the power of 4 minus 16)
one divide by (x to the power of four minus sixteen)
1/(x4-16)
1/x4-16
1/(x⁴-16)
1/x^4-16
1 divide by (x^4-16)
1/(x^4-16)dx
Similar expressions
1/(x^4+16)

Integral of 1/(x^4-16) dx

The solution

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 oo           
  /           
 |            
 |     1      
 |  ------- dx
 |   4        
 |  x  - 16   
 |            
/             
2

$$\int\limits_{2}^{\infty} \frac{1}{x^{4} - 16}\, dx$$

Integral(1/(x^4 - 16), (x, 2, oo))

Detail solution

Rewrite the integrand:

$\frac{1}{x^{4} - 16} = - \frac{1}{8 \left(x^{2} + 4\right)} - \frac{1}{32 \left(x + 2\right)} + \frac{1}{32 \left(x - 2\right)}$
Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{1}{8 \left(x^{2} + 4\right)}\right)\, dx = - \frac{\int \frac{1}{x^{2} + 4}\, dx}{8}$
  So, the result is: $- \frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{16}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{1}{32 \left(x + 2\right)}\right)\, dx = - \frac{\int \frac{1}{x + 2}\, dx}{32}$
  1. Let $u = x + 2$ .
    
    Then let $du = dx$ and substitute $du$ :
    
    $\int \frac{1}{u}\, du$
    1. 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: $- \frac{\log{\left(x + 2 \right)}}{32}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{1}{32 \left(x - 2\right)}\, dx = \frac{\int \frac{1}{x - 2}\, dx}{32}$
  1. Let $u = x - 2$ .
    
    Then let $du = dx$ and substitute $du$ :
    
    $\int \frac{1}{u}\, du$
    1. 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: $\frac{\log{\left(x - 2 \right)}}{32}$
The result is: $\frac{\log{\left(x - 2 \right)}}{32} - \frac{\log{\left(x + 2 \right)}}{32} - \frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{16}$
Add the constant of integration:

$\frac{\log{\left(x - 2 \right)}}{32} - \frac{\log{\left(x + 2 \right)}}{32} - \frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{16}+ \mathrm{constant}$

The answer is:

$\frac{\log{\left(x - 2 \right)}}{32} - \frac{\log{\left(x + 2 \right)}}{32} - \frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{16}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                     /x\                           
 |                  atan|-|                           
 |    1                 \2/   log(2 + x)   log(-2 + x)
 | ------- dx = C - ------- - ---------- + -----------
 |  4                  16         32            32    
 | x  - 16                                            
 |                                                    
/

$$\int \frac{1}{x^{4} - 16}\, dx = C + \frac{\log{\left(x - 2 \right)}}{32} - \frac{\log{\left(x + 2 \right)}}{32} - \frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{16}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

oo

$$\infty$$

oo

$$\infty$$

oo

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

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Integral of 1/(x^4-16) dx

Limits of integration:

The graph:

Piecewise:

The solution