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Solving integrals/ 2*dx/((2*x+1)^(2)*(x-1))

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

The solution

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

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

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

Detail solution

The integral of a constant times a function is the constant times the integral of the function:

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

$\frac{2 \left(\left(2 x + 1\right) \left(\log{\left(x - 1 \right)} - \log{\left(2 x + 1 \right)}\right) + 3\right)}{9 \left(2 x + 1\right)}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

      2*pi*I
-oo - ------
        9

$$-\infty - \frac{2 i \pi}{9}$$

      2*pi*I
-oo - ------
        9

$$-\infty - \frac{2 i \pi}{9}$$

-oo - 2*pi*i/9

Numerical answer [src]

-10.4865709055306

-10.4865709055306

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

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How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

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