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(x^ three - one)/(4x^ three -x)
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Similar expressions
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Solving integrals/ (x^3-1)/(4x^3-x)

Integral of (x^3-1)/(4x^3-x) dx

The solution

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

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

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{x^{3} - 1}{4 x^{3} - x} = \frac{1}{4} - \frac{9}{8 \left(2 x + 1\right)} - \frac{7}{8 \left(2 x - 1\right)} + \frac{1}{x}$
2. Integrate term-by-term:
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{4}\, dx = \frac{x}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{9}{8 \left(2 x + 1\right)}\right)\, dx = - \frac{9 \int \frac{1}{2 x + 1}\, dx}{8}$
    1. Let $u = 2 x + 1$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{2 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(2 x + 1 \right)}}{2}$
    So, the result is: $- \frac{9 \log{\left(2 x + 1 \right)}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{7}{8 \left(2 x - 1\right)}\right)\, dx = - \frac{7 \int \frac{1}{2 x - 1}\, dx}{8}$
    1. Let $u = 2 x - 1$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{2 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(2 x - 1 \right)}}{2}$
    So, the result is: $- \frac{7 \log{\left(2 x - 1 \right)}}{16}$
  1. The integral of $\frac{1}{x}$ is $\log{\left(x \right)}$ .
  The result is: $\frac{x}{4} + \log{\left(x \right)} - \frac{7 \log{\left(2 x - 1 \right)}}{16} - \frac{9 \log{\left(2 x + 1 \right)}}{16}$
Method #2
1. Rewrite the integrand:
  
  $\frac{x^{3} - 1}{4 x^{3} - x} = \frac{x^{3}}{4 x^{3} - x} - \frac{1}{4 x^{3} - x}$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\frac{x^{3}}{4 x^{3} - x} = \frac{1}{4} - \frac{1}{8 \left(2 x + 1\right)} + \frac{1}{8 \left(2 x - 1\right)}$
  2. Integrate term-by-term:
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, dx = \frac{x}{4}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{8 \left(2 x + 1\right)}\right)\, dx = - \frac{\int \frac{1}{2 x + 1}\, dx}{8}$
      
      Let $u = 2 x + 1$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{2 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(2 x + 1 \right)}}{2}$
      
      So, the result is: $- \frac{\log{\left(2 x + 1 \right)}}{16}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{8 \left(2 x - 1\right)}\, dx = \frac{\int \frac{1}{2 x - 1}\, dx}{8}$
      
      Let $u = 2 x - 1$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{2 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(2 x - 1 \right)}}{2}$
      
      So, the result is: $\frac{\log{\left(2 x - 1 \right)}}{16}$
    The result is: $\frac{x}{4} + \frac{\log{\left(2 x - 1 \right)}}{16} - \frac{\log{\left(2 x + 1 \right)}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{4 x^{3} - x}\right)\, dx = - \int \frac{1}{4 x^{3} - x}\, dx$
    1. Rewrite the integrand:
      
      $\frac{1}{4 x^{3} - x} = \frac{1}{2 x + 1} + \frac{1}{2 x - 1} - \frac{1}{x}$
    2. Integrate term-by-term:
      
      Let $u = 2 x + 1$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{2 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(2 x + 1 \right)}}{2}$
      
      Let $u = 2 x - 1$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{2 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(2 x - 1 \right)}}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{x}\right)\, dx = - \int \frac{1}{x}\, dx$
      
      The integral of $\frac{1}{x}$ is $\log{\left(x \right)}$ .
      
      So, the result is: $- \log{\left(x \right)}$
      
      The result is: $- \log{\left(x \right)} + \frac{\log{\left(2 x - 1 \right)}}{2} + \frac{\log{\left(2 x + 1 \right)}}{2}$
    So, the result is: $\log{\left(x \right)} - \frac{\log{\left(2 x - 1 \right)}}{2} - \frac{\log{\left(2 x + 1 \right)}}{2}$
  The result is: $\frac{x}{4} + \log{\left(x \right)} - \frac{7 \log{\left(2 x - 1 \right)}}{16} - \frac{9 \log{\left(2 x + 1 \right)}}{16}$
Add the constant of integration:

$\frac{x}{4} + \log{\left(x \right)} - \frac{7 \log{\left(2 x - 1 \right)}}{16} - \frac{9 \log{\left(2 x + 1 \right)}}{16}+ \mathrm{constant}$

The answer is:

$\frac{x}{4} + \log{\left(x \right)} - \frac{7 \log{\left(2 x - 1 \right)}}{16} - \frac{9 \log{\left(2 x + 1 \right)}}{16}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                               
 |                                                                
 |   3                                                            
 |  x  - 1           9*log(1 + 2*x)   7*log(-1 + 2*x)   x         
 | -------- dx = C - -------------- - --------------- + - + log(x)
 |    3                    16                16         4         
 | 4*x  - x                                                       
 |                                                                
/

$$\int \frac{x^{3} - 1}{4 x^{3} - x}\, dx = C + \frac{x}{4} + \log{\left(x \right)} - \frac{7 \log{\left(2 x - 1 \right)}}{16} - \frac{9 \log{\left(2 x + 1 \right)}}{16}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

nan

$$\text{NaN}$$

nan

$$\text{NaN}$$

nan

The graph

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

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Identical expressions

Similar expressions

Integral of (x^3-1)/(4x^3-x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2