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Solving integrals/ (x^2-1)/(2x-1)

Integral of (x^2-1)/(2x-1) dx

The solution

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

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

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

Detail solution

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

$\frac{x^{2}}{4} + \frac{x}{4} - \frac{3 \log{\left(2 x - 1 \right)}}{8}+ \mathrm{constant}$

The answer is:

\frac{x^{2}}{4} + \frac{x}{4} - \frac{3 \log{\left(2 x - 1 \right)}}{8}+ \mathrm{constant}

The answer (Indefinite) [src]

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

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

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.

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (x^2-1)/(2x-1) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2