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Graphing y =:
(2x-3)/(x+1)
Identical expressions
(2x- three)/(x+ one)
(2x minus 3) divide by (x plus 1)
(2x minus three) divide by (x plus one)
2x-3/x+1
(2x-3) divide by (x+1)
(2x-3)/(x+1)dx
Similar expressions
(2x^2+2x-3)/(x+1)
(2x+3)/(x+1)
(2x-3)/(x-1)

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

The solution

You have entered [src]

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

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

Integral((2*x - 3)/(x + 1), (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 $du$ :
  
  $\int \frac{u - 3}{u + 2}\, du$
  1. Let $u = u + 2$ .
    
    Then let $du = du$ and substitute $du$ :
    
    $\int \frac{u - 5}{u}\, du$
    1. Rewrite the integrand:
      
      $\frac{u - 5}{u} = 1 - \frac{5}{u}$
    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:
      
      $\int \left(- \frac{5}{u}\right)\, du = - 5 \int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $- 5 \log{\left(u \right)}$
      
      The result is: $u - 5 \log{\left(u \right)}$
    Now substitute $u$ back in:
    
    $u - 5 \log{\left(u + 2 \right)} + 2$
  Now substitute $u$ back in:
  
  $2 x - 5 \log{\left(2 x + 2 \right)} + 2$
Method #2
1. Rewrite the integrand:
  
  $\frac{2 x - 3}{x + 1} = 2 - \frac{5}{x + 1}$
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:
    
    $\int \left(- \frac{5}{x + 1}\right)\, dx = - 5 \int \frac{1}{x + 1}\, dx$
    1. 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: $- 5 \log{\left(x + 1 \right)}$
  The result is: $2 x - 5 \log{\left(x + 1 \right)}$
Method #3
1. Rewrite the integrand:
  
  $\frac{2 x - 3}{x + 1} = \frac{2 x}{x + 1} - \frac{3}{x + 1}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{2 x}{x + 1}\, dx = 2 \int \frac{x}{x + 1}\, dx$
    1. Rewrite the integrand:
      
      $\frac{x}{x + 1} = 1 - \frac{1}{x + 1}$
    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:
      
      $\int \left(- \frac{1}{x + 1}\right)\, dx = - \int \frac{1}{x + 1}\, dx$
      
      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: $- \log{\left(x + 1 \right)}$
      
      The result is: $x - \log{\left(x + 1 \right)}$
    So, the result is: $2 x - 2 \log{\left(x + 1 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{3}{x + 1}\right)\, dx = - 3 \int \frac{1}{x + 1}\, dx$
    1. 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: $- 3 \log{\left(x + 1 \right)}$
  The result is: $2 x - 3 \log{\left(x + 1 \right)} - 2 \log{\left(x + 1 \right)}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

2 - 5*log(2)

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

2 - 5*log(2)

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

2 - 5*log(2)

Numerical answer [src]

-1.46573590279973

-1.46573590279973

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3