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Identical expressions
(x^ three)/(nine -4x^ eight)
(x cubed ) divide by (9 minus 4x to the power of 8)
(x to the power of three) divide by (nine minus 4x to the power of eight)
(x3)/(9-4x8)
x3/9-4x8
(x³)/(9-4x⁸)
(x to the power of 3)/(9-4x to the power of 8)
x^3/9-4x^8
(x^3) divide by (9-4x^8)
(x^3)/(9-4x^8)dx
Similar expressions
(x^3)/(9+4x^8)

Solving integrals/ (x^3)/(9-4x^8)

Integral of (x^3)/(9-4x^8) dx

The solution

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  1            
  /            
 |             
 |      3      
 |     x       
 |  -------- dx
 |         8   
 |  9 - 4*x    
 |             
/              
0

$$\int\limits_{0}^{1} \frac{x^{3}}{9 - 4 x^{8}}\, dx$$

Integral(x^3/(9 - 4*x^8), (x, 0, 1))

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 $- du$ :
  
  $\int \frac{u}{8 u^{4} - 18}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{u}{8 u^{4} - 18}\right)\, du = - \int \frac{u}{8 u^{4} - 18}\, du$
    1. Let $u = u^{2}$ .
      
      Then let $du = 2 u du$ and substitute $du$ :
      
      $\int \frac{1}{16 u^{2} - 36}\, du$
      
      Rewrite the integrand:
      
      $\frac{1}{16 u^{2} - 36} = \frac{- \frac{1}{u + \frac{3}{2}} + \frac{1}{u - \frac{3}{2}}}{48}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{- \frac{1}{u + \frac{3}{2}} + \frac{1}{u - \frac{3}{2}}}{48}\, du = \frac{\int \left(- \frac{1}{u + \frac{3}{2}} + \frac{1}{u - \frac{3}{2}}\right)\, du}{48}$
      
      Integrate term-by-term:
      
      The integral of $\frac{1}{u - \frac{3}{2}}$ is $\log{\left(u - \frac{3}{2} \right)}$ .
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u + \frac{3}{2}}\right)\, du = - \int \frac{1}{u + \frac{3}{2}}\, du$
      
      The integral of $\frac{1}{u + \frac{3}{2}}$ is $\log{\left(u + \frac{3}{2} \right)}$ .
      
      So, the result is: $- \log{\left(u + \frac{3}{2} \right)}$
      
      The result is: $\log{\left(u - \frac{3}{2} \right)} - \log{\left(u + \frac{3}{2} \right)}$
      
      So, the result is: $\frac{\log{\left(u - \frac{3}{2} \right)}}{48} - \frac{\log{\left(u + \frac{3}{2} \right)}}{48}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(u^{2} - \frac{3}{2} \right)}}{48} - \frac{\log{\left(u^{2} + \frac{3}{2} \right)}}{48}$
    So, the result is: $- \frac{\log{\left(u^{2} - \frac{3}{2} \right)}}{48} + \frac{\log{\left(u^{2} + \frac{3}{2} \right)}}{48}$
  Now substitute $u$ back in:
  
  $- \frac{\log{\left(x^{4} - \frac{3}{2} \right)}}{48} + \frac{\log{\left(x^{4} + \frac{3}{2} \right)}}{48}$
Method #2
1. Rewrite the integrand:
  
  $\frac{x^{3}}{9 - 4 x^{8}} = \frac{x^{3}}{6 \cdot \left(2 x^{4} + 3\right)} - \frac{x^{3}}{6 \cdot \left(2 x^{4} - 3\right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{x^{3}}{6 \cdot \left(2 x^{4} + 3\right)}\, dx = \frac{\int \frac{x^{3}}{2 x^{4} + 3}\, dx}{6}$
    1. Let $u = 2 x^{4} + 3$ .
      
      Then let $du = 8 x^{3} dx$ and substitute $\frac{du}{8}$ :
      
      $\int \frac{1}{64 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{8 u}\, du = \frac{\int \frac{1}{u}\, du}{8}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{8}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(2 x^{4} + 3 \right)}}{8}$
    So, the result is: $\frac{\log{\left(2 x^{4} + 3 \right)}}{48}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{x^{3}}{6 \cdot \left(2 x^{4} - 3\right)}\right)\, dx = - \frac{\int \frac{x^{3}}{2 x^{4} - 3}\, dx}{6}$
    1. Let $u = 2 x^{4} - 3$ .
      
      Then let $du = 8 x^{3} dx$ and substitute $\frac{du}{8}$ :
      
      $\int \frac{1}{64 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{8 u}\, du = \frac{\int \frac{1}{u}\, du}{8}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{8}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(2 x^{4} - 3 \right)}}{8}$
    So, the result is: $- \frac{\log{\left(2 x^{4} - 3 \right)}}{48}$
  The result is: $- \frac{\log{\left(2 x^{4} - 3 \right)}}{48} + \frac{\log{\left(2 x^{4} + 3 \right)}}{48}$
Method #3
1. Rewrite the integrand:
  
  $\frac{x^{3}}{9 - 4 x^{8}} = - \frac{x^{3}}{4 x^{8} - 9}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{x^{3}}{4 x^{8} - 9}\right)\, dx = - \int \frac{x^{3}}{4 x^{8} - 9}\, dx$
  1. Let $u = x^{2}$ .
    
    Then let $du = 2 x dx$ and substitute $du$ :
    
    $\int \frac{u}{8 u^{4} - 18}\, du$
    1. Let $u = u^{2}$ .
      
      Then let $du = 2 u du$ and substitute $du$ :
      
      $\int \frac{1}{16 u^{2} - 36}\, du$
      
      Rewrite the integrand:
      
      $\frac{1}{16 u^{2} - 36} = \frac{- \frac{1}{u + \frac{3}{2}} + \frac{1}{u - \frac{3}{2}}}{48}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{- \frac{1}{u + \frac{3}{2}} + \frac{1}{u - \frac{3}{2}}}{48}\, du = \frac{\int \left(- \frac{1}{u + \frac{3}{2}} + \frac{1}{u - \frac{3}{2}}\right)\, du}{48}$
      
      Integrate term-by-term:
      
      The integral of $\frac{1}{u - \frac{3}{2}}$ is $\log{\left(u - \frac{3}{2} \right)}$ .
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u + \frac{3}{2}}\right)\, du = - \int \frac{1}{u + \frac{3}{2}}\, du$
      
      The integral of $\frac{1}{u + \frac{3}{2}}$ is $\log{\left(u + \frac{3}{2} \right)}$ .
      
      So, the result is: $- \log{\left(u + \frac{3}{2} \right)}$
      
      The result is: $\log{\left(u - \frac{3}{2} \right)} - \log{\left(u + \frac{3}{2} \right)}$
      
      So, the result is: $\frac{\log{\left(u - \frac{3}{2} \right)}}{48} - \frac{\log{\left(u + \frac{3}{2} \right)}}{48}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(u^{2} - \frac{3}{2} \right)}}{48} - \frac{\log{\left(u^{2} + \frac{3}{2} \right)}}{48}$
    Now substitute $u$ back in:
    
    $\frac{\log{\left(x^{4} - \frac{3}{2} \right)}}{48} - \frac{\log{\left(x^{4} + \frac{3}{2} \right)}}{48}$
  So, the result is: $- \frac{\log{\left(x^{4} - \frac{3}{2} \right)}}{48} + \frac{\log{\left(x^{4} + \frac{3}{2} \right)}}{48}$
Add the constant of integration:

$- \frac{\log{\left(x^{4} - \frac{3}{2} \right)}}{48} + \frac{\log{\left(x^{4} + \frac{3}{2} \right)}}{48}+ \mathrm{constant}$

The answer is:

$- \frac{\log{\left(x^{4} - \frac{3}{2} \right)}}{48} + \frac{\log{\left(x^{4} + \frac{3}{2} \right)}}{48}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                             
 |                      /  3    4\      /3    4\
 |     3             log|- - + x |   log|- + x |
 |    x                 \  2     /      \2     /
 | -------- dx = C - ------------- + -----------
 |        8                48             48    
 | 9 - 4*x                                      
 |                                              
/

$${{\log \left(2\,x^4+3\right)}\over{48}}-{{\log \left(2\,x^4-3 \right)}\over{48}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

log(2)   log(5/2)
------ + --------
  48        48

$${{\log 5}\over{48}}$$

log(2)   log(5/2)
------ + --------
  48        48

$$\frac{\log{\left(2 \right)}}{48} + \frac{\log{\left(\frac{5}{2} \right)}}{48}$$

Упростить

Numerical answer [src]

0.0335299565090437

0.0335299565090437

The graph

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

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How to use it?

Identical expressions

Similar expressions

Integral of (x^3)/(9-4x^8) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3