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Identical expressions
x/(2x+ one)(2x- three)
x divide by (2x plus 1)(2x minus 3)
x divide by (2x plus one)(2x minus three)
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x divide by (2x+1)(2x-3)
x/(2x+1)(2x-3)dx
Similar expressions
x/(2x-1)(2x-3)
x/(2x+1)(2x+3)

Integral of x/(2x+1)(2x-3) dx

The solution

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

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

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

Detail solution

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

$\frac{x^{2}}{2} - 2 x + \log{\left(2 x + 1 \right)}+ \mathrm{constant}$

The answer is:

$\frac{x^{2}}{2} - 2 x + \log{\left(2 x + 1 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                  
 |                             2                     
 |    x                       x                      
 | -------*(2*x - 3) dx = C + -- - 2*x + log(1 + 2*x)
 | 2*x + 1                    2                      
 |                                                   
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

-3/2 + log(3)

$$- \frac{3}{2} + \log{\left(3 \right)}$$

-3/2 + log(3)

$$- \frac{3}{2} + \log{\left(3 \right)}$$

-3/2 + log(3)

Numerical answer [src]

-0.40138771133189

-0.40138771133189

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

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Identical expressions

Similar expressions

Integral of x/(2x+1)(2x-3) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3