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Identical expressions
((4x+ two)/(2x- one))dx
((4x plus 2) divide by (2x minus 1))dx
((4x plus two) divide by (2x minus one))dx
4x+2/2x-1dx
((4x+2) divide by (2x-1))dx
Similar expressions
((4x+2)/(2x+1))dx
((4x-2)/(2x-1))dx

Integral of ((4x+2)/(2x-1))dx dx

The solution

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  2           
  /           
 |            
 |  4*x + 2   
 |  ------- dx
 |  2*x - 1   
 |            
/             
1

$$\int\limits_{1}^{2} \frac{4 x + 2}{2 x - 1}\, dx$$

Integral((4*x + 2)/(2*x - 1), (x, 1, 2))

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

2 + 2*log(3)

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

2 + 2*log(3)

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

2 + 2*log(3)

Numerical answer [src]

4.19722457733622

4.19722457733622

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

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Identical expressions

Similar expressions

Integral of ((4x+2)/(2x-1))dx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3