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(x³- one)/(2x+ one)
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Similar expressions
(x³-1)/(2x-1)
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Solving integrals/ (x³-1)/(2x+1)

Integral of (x³-1)/(2x+1) dx

The solution

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

$$\int\limits_{0}^{1} \frac{x^{3} - 1}{2 x + 1}\, dx$$

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{x^{3} - 1}{2 x + 1} = \frac{x^{2}}{2} - \frac{x}{4} + \frac{1}{8} - \frac{9}{8 \left(2 x + 1\right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{x^{2}}{2}\, dx = \frac{\int x^{2}\, dx}{2}$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{2}\, dx = \frac{x^{3}}{3}$
    So, the result is: $\frac{x^{3}}{6}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{x}{4}\right)\, dx = - \frac{\int x\, dx}{4}$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
    So, the result is: $- \frac{x^{2}}{8}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{8}\, dx = \frac{x}{8}$
  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: $- \frac{9 \log{\left(2 x + 1 \right)}}{16}$
  The result is: $\frac{x^{3}}{6} - \frac{x^{2}}{8} + \frac{x}{8} - \frac{9 \log{\left(2 x + 1 \right)}}{16}$
Method #2
1. Rewrite the integrand:
  
  $\frac{x^{3} - 1}{2 x + 1} = \frac{x^{3}}{2 x + 1} - \frac{1}{2 x + 1}$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\frac{x^{3}}{2 x + 1} = \frac{x^{2}}{2} - \frac{x}{4} + \frac{1}{8} - \frac{1}{8 \left(2 x + 1\right)}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{x^{2}}{2}\, dx = \frac{\int x^{2}\, dx}{2}$
      
      The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{2}\, dx = \frac{x^{3}}{3}$
      
      So, the result is: $\frac{x^{3}}{6}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{x}{4}\right)\, dx = - \frac{\int x\, dx}{4}$
      
      The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
      
      So, the result is: $- \frac{x^{2}}{8}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{8}\, dx = \frac{x}{8}$
    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: $- \frac{\log{\left(2 x + 1 \right)}}{16}$
    The result is: $\frac{x^{3}}{6} - \frac{x^{2}}{8} + \frac{x}{8} - \frac{\log{\left(2 x + 1 \right)}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{2 x + 1}\right)\, dx = - \int \frac{1}{2 x + 1}\, dx$
    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: $- \frac{\log{\left(2 x + 1 \right)}}{2}$
  The result is: $\frac{x^{3}}{6} - \frac{x^{2}}{8} + \frac{x}{8} - \frac{\log{\left(2 x + 1 \right)}}{16} - \frac{\log{\left(2 x + 1 \right)}}{2}$
Add the constant of integration:

$\frac{x^{3}}{6} - \frac{x^{2}}{8} + \frac{x}{8} - \frac{9 \log{\left(2 x + 1 \right)}}{16}+ \mathrm{constant}$

The answer is:

$\frac{x^{3}}{6} - \frac{x^{2}}{8} + \frac{x}{8} - \frac{9 \log{\left(2 x + 1 \right)}}{16}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                             
 |                                              
 |   3                                2    3    
 |  x  - 1          9*log(1 + 2*x)   x    x    x
 | ------- dx = C - -------------- - -- + -- + -
 | 2*x + 1                16         8    6    8
 |                                              
/

$$\int \frac{x^{3} - 1}{2 x + 1}\, dx = C + \frac{x^{3}}{6} - \frac{x^{2}}{8} + \frac{x}{8} - \frac{9 \log{\left(2 x + 1 \right)}}{16}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

1   9*log(3)
- - --------
6      16

$$\frac{1}{6} - \frac{9 \log{\left(3 \right)}}{16}$$

1   9*log(3)
- - --------
6      16

$$\frac{1}{6} - \frac{9 \log{\left(3 \right)}}{16}$$

1/6 - 9*log(3)/16

Numerical answer [src]

-0.451302745709145

-0.451302745709145

The graph

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

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Identical expressions

Similar expressions

Integral of (x³-1)/(2x+1) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2