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Graphing y =:
(2x-1)/(x+4)
Identical expressions
(2x- one)/(x+ four)
(2x minus 1) divide by (x plus 4)
(2x minus one) divide by (x plus four)
2x-1/x+4
(2x-1) divide by (x+4)
(2x-1)/(x+4)dx
Similar expressions
(2x-1)/(x-4)
(2x+1)/(x+4)

Solving integrals/ (2x-1)/(x+4)

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

The solution

You have entered [src]

  1           
  /           
 |            
 |  2*x - 1   
 |  ------- dx
 |   x + 4    
 |            
/             
0

$$\int\limits_{0}^{1} \frac{2 x - 1}{x + 4}\, dx$$

Integral((2*x - 1)/(x + 4), (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 $du$ :
  
  $\int \frac{u - 1}{u + 8}\, du$
  1. Let $u = u + 8$ .
    
    Then let $du = du$ and substitute $du$ :
    
    $\int \frac{u - 9}{u}\, du$
    1. Rewrite the integrand:
      
      $\frac{u - 9}{u} = 1 - \frac{9}{u}$
    2. Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{9}{u}\right)\, du = - 9 \int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $- 9 \log{\left(u \right)}$
      
      The result is: $u - 9 \log{\left(u \right)}$
    Now substitute $u$ back in:
    
    $u - 9 \log{\left(u + 8 \right)} + 8$
  Now substitute $u$ back in:
  
  $2 x - 9 \log{\left(2 x + 8 \right)} + 8$
Method #2
1. Rewrite the integrand:
  
  $\frac{2 x - 1}{x + 4} = 2 - \frac{9}{x + 4}$
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:
    
    $\int \left(- \frac{9}{x + 4}\right)\, dx = - 9 \int \frac{1}{x + 4}\, dx$
    1. Let $u = x + 4$ .
      
      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:
      
      $\log{\left(x + 4 \right)}$
    So, the result is: $- 9 \log{\left(x + 4 \right)}$
  The result is: $2 x - 9 \log{\left(x + 4 \right)}$
Method #3
1. Rewrite the integrand:
  
  $\frac{2 x - 1}{x + 4} = \frac{2 x}{x + 4} - \frac{1}{x + 4}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{2 x}{x + 4}\, dx = 2 \int \frac{x}{x + 4}\, dx$
    1. Rewrite the integrand:
      
      $\frac{x}{x + 4} = 1 - \frac{4}{x + 4}$
    2. Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{4}{x + 4}\right)\, dx = - 4 \int \frac{1}{x + 4}\, dx$
      
      Let $u = x + 4$ .
      
      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:
      
      $\log{\left(x + 4 \right)}$
      
      So, the result is: $- 4 \log{\left(x + 4 \right)}$
      
      The result is: $x - 4 \log{\left(x + 4 \right)}$
    So, the result is: $2 x - 8 \log{\left(x + 4 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{x + 4}\right)\, dx = - \int \frac{1}{x + 4}\, dx$
    1. Let $u = x + 4$ .
      
      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:
      
      $\log{\left(x + 4 \right)}$
    So, the result is: $- \log{\left(x + 4 \right)}$
  The result is: $2 x - \log{\left(x + 4 \right)} - 8 \log{\left(x + 4 \right)}$
Add the constant of integration:

$2 x - 9 \log{\left(2 x + 8 \right)} + 8+ \mathrm{constant}$

The answer is:

$2 x - 9 \log{\left(2 x + 8 \right)} + 8+ \mathrm{constant}$

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

2 - 9*log(5) + 9*log(4)

$$- 9 \log{\left(5 \right)} + 2 + 9 \log{\left(4 \right)}$$

2 - 9*log(5) + 9*log(4)

$$- 9 \log{\left(5 \right)} + 2 + 9 \log{\left(4 \right)}$$

2 - 9*log(5) + 9*log(4)

Numerical answer [src]

-0.0082919618278878

-0.0082919618278878

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3