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

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

The solution

You have entered [src]

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

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

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{x - 1}{2 x + 1} \left(x - 3\right) = \frac{x}{2} - \frac{9}{4} + \frac{21}{4 \left(2 x + 1\right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{x}{2}\, dx = \frac{\int x\, dx}{2}$
    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: $\frac{x^{2}}{4}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \left(- \frac{9}{4}\right)\, dx = - \frac{9 x}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{21}{4 \left(2 x + 1\right)}\, dx = \frac{21 \int \frac{1}{2 x + 1}\, dx}{4}$
    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: $\frac{21 \log{\left(2 x + 1 \right)}}{8}$
  The result is: $\frac{x^{2}}{4} - \frac{9 x}{4} + \frac{21 \log{\left(2 x + 1 \right)}}{8}$
Method #2
1. Rewrite the integrand:
  
  $\frac{x - 1}{2 x + 1} \left(x - 3\right) = \frac{x^{2} - 4 x + 3}{2 x + 1}$
2. Rewrite the integrand:
  
  $\frac{x^{2} - 4 x + 3}{2 x + 1} = \frac{x}{2} - \frac{9}{4} + \frac{21}{4 \left(2 x + 1\right)}$
3. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{x}{2}\, dx = \frac{\int x\, dx}{2}$
    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: $\frac{x^{2}}{4}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \left(- \frac{9}{4}\right)\, dx = - \frac{9 x}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{21}{4 \left(2 x + 1\right)}\, dx = \frac{21 \int \frac{1}{2 x + 1}\, dx}{4}$
    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: $\frac{21 \log{\left(2 x + 1 \right)}}{8}$
  The result is: $\frac{x^{2}}{4} - \frac{9 x}{4} + \frac{21 \log{\left(2 x + 1 \right)}}{8}$
Method #3
1. Rewrite the integrand:
  
  $\frac{x - 1}{2 x + 1} \left(x - 3\right) = \frac{x^{2}}{2 x + 1} - \frac{4 x}{2 x + 1} + \frac{3}{2 x + 1}$
2. Integrate term-by-term:
  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:
    1. 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: $\frac{x^{2}}{4}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \left(- \frac{1}{4}\right)\, dx = - \frac{x}{4}$
    1. 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: $\frac{x^{2}}{4} - \frac{x}{4} + \frac{\log{\left(2 x + 1 \right)}}{8}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{4 x}{2 x + 1}\right)\, dx = - 4 \int \frac{x}{2 x + 1}\, dx$
    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: $- 2 x + \log{\left(2 x + 1 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{3}{2 x + 1}\, dx = 3 \int \frac{1}{2 x + 1}\, dx$
    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: $\frac{3 \log{\left(2 x + 1 \right)}}{2}$
  The result is: $\frac{x^{2}}{4} - \frac{9 x}{4} + \frac{3 \log{\left(2 x + 1 \right)}}{2} + \frac{9 \log{\left(2 x + 1 \right)}}{8}$
Add the constant of integration:

$\frac{x^{2}}{4} - \frac{9 x}{4} + \frac{21 \log{\left(2 x + 1 \right)}}{8}+ \mathrm{constant}$

The answer is:

$\frac{x^{2}}{4} - \frac{9 x}{4} + \frac{21 \log{\left(2 x + 1 \right)}}{8}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                   
 |                                 2                  
 |  x - 1                   9*x   x    21*log(1 + 2*x)
 | -------*(x - 3) dx = C - --- + -- + ---------------
 | 2*x + 1                   4    4           8       
 |                                                    
/

$$\int \frac{x - 1}{2 x + 1} \left(x - 3\right)\, dx = C + \frac{x^{2}}{4} - \frac{9 x}{4} + \frac{21 \log{\left(2 x + 1 \right)}}{8}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

     21*log(3)
-2 + ---------
         8

$$-2 + \frac{21 \log{\left(3 \right)}}{8}$$

     21*log(3)
-2 + ---------
         8

$$-2 + \frac{21 \log{\left(3 \right)}}{8}$$

-2 + 21*log(3)/8

Numerical answer [src]

0.883857257753788

0.883857257753788

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3