Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of e^(-x)
Integral of -sin(x)
Integral of cos(x/2)
Integral of x-2
Identical expressions
one /(2x- one)(2x+ one)(2x+ three)
1 divide by (2x minus 1)(2x plus 1)(2x plus 3)
one divide by (2x minus one)(2x plus one)(2x plus three)
1/2x-12x+12x+3
1 divide by (2x-1)(2x+1)(2x+3)
1/(2x-1)(2x+1)(2x+3)dx
Similar expressions
1/(2x-1)(2x-1)(2x+3)
1/(2x-1)(2x+1)(2x-3)
1/(2x+1)(2x+1)(2x+3)

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

The solution

You have entered [src]

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

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

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = 2 x$ .
  
  Then let $du = 2 dx$ and substitute $du$ :
  
  $\int \frac{u^{2} + 4 u + 3}{2 u - 2}\, du$
  1. Rewrite the integrand:
    
    $\frac{u^{2} + 4 u + 3}{2 u - 2} = \frac{u}{2} + \frac{5}{2} + \frac{4}{u - 1}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u}{2}\, du = \frac{\int u\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $\frac{u^{2}}{4}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{5}{2}\, du = \frac{5 u}{2}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{4}{u - 1}\, du = 4 \int \frac{1}{u - 1}\, du$
      
      Let $u = u - 1$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(u - 1 \right)}$
      
      So, the result is: $4 \log{\left(u - 1 \right)}$
    The result is: $\frac{u^{2}}{4} + \frac{5 u}{2} + 4 \log{\left(u - 1 \right)}$
  Now substitute $u$ back in:
  
  $x^{2} + 5 x + 4 \log{\left(2 x - 1 \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{2 x + 1}{2 x - 1} \left(2 x + 3\right) = 2 x + 5 + \frac{8}{2 x - 1}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 x\, dx = 2 \int x\, dx$
    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: $x^{2}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 5\, dx = 5 x$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{8}{2 x - 1}\, dx = 8 \int \frac{1}{2 x - 1}\, dx$
    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: $4 \log{\left(2 x - 1 \right)}$
  The result is: $x^{2} + 5 x + 4 \log{\left(2 x - 1 \right)}$
Method #3
1. Rewrite the integrand:
  
  $\frac{2 x + 1}{2 x - 1} \left(2 x + 3\right) = \frac{4 x^{2} + 8 x + 3}{2 x - 1}$
2. Rewrite the integrand:
  
  $\frac{4 x^{2} + 8 x + 3}{2 x - 1} = 2 x + 5 + \frac{8}{2 x - 1}$
3. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 x\, dx = 2 \int x\, dx$
    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: $x^{2}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 5\, dx = 5 x$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{8}{2 x - 1}\, dx = 8 \int \frac{1}{2 x - 1}\, dx$
    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: $4 \log{\left(2 x - 1 \right)}$
  The result is: $x^{2} + 5 x + 4 \log{\left(2 x - 1 \right)}$
Method #4
1. Rewrite the integrand:
  
  $\frac{2 x + 1}{2 x - 1} \left(2 x + 3\right) = \frac{4 x^{2}}{2 x - 1} + \frac{8 x}{2 x - 1} + \frac{3}{2 x - 1}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{4 x^{2}}{2 x - 1}\, dx = 4 \int \frac{x^{2}}{2 x - 1}\, dx$
    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:
      
      $\int \frac{1}{4}\, dx = \frac{x}{4}$
      
      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: $x^{2} + x + \frac{\log{\left(2 x - 1 \right)}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{8 x}{2 x - 1}\, dx = 8 \int \frac{x}{2 x - 1}\, dx$
    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:
      
      $\int \frac{1}{2 \left(2 x - 1\right)}\, dx = \frac{\int \frac{1}{2 x - 1}\, dx}{2}$
      
      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: $4 x + 2 \log{\left(2 x - 1 \right)}$
  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: $x^{2} + 5 x + \frac{5 \log{\left(2 x - 1 \right)}}{2} + \frac{3 \log{\left(2 x - 1 \right)}}{2}$
Add the constant of integration:

$x^{2} + 5 x + 4 \log{\left(2 x - 1 \right)}+ \mathrm{constant}$

The answer is:

$x^{2} + 5 x + 4 \log{\left(2 x - 1 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                     
 |                                                      
 | 2*x + 1                     2                        
 | -------*(2*x + 3) dx = C + x  + 4*log(-1 + 2*x) + 5*x
 | 2*x - 1                                              
 |                                                      
/

$$\int \frac{2 x + 1}{2 x - 1} \left(2 x + 3\right)\, dx = C + x^{2} + 5 x + 4 \log{\left(2 x - 1 \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

nan

$$\text{NaN}$$

nan

$$\text{NaN}$$

nan

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

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4