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Integral of d{x}:
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Identical expressions
(three x+ two)/(x-3)
(3x plus 2) divide by (x minus 3)
(three x plus two) divide by (x minus 3)
3x+2/x-3
(3x+2) divide by (x-3)
(3x+2)/(x-3)dx
Similar expressions
(3x-2)/(x-3)
(3x+2)/(x+3)

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

The solution

You have entered [src]

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

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

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = 3 x$ .
  
  Then let $du = 3 dx$ and substitute $du$ :
  
  $\int \frac{u + 2}{u - 9}\, du$
  1. Let $u = u - 9$ .
    
    Then let $du = du$ and substitute $du$ :
    
    $\int \frac{u + 11}{u}\, du$
    1. Rewrite the integrand:
      
      $\frac{u + 11}{u} = 1 + \frac{11}{u}$
    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:
      
      $\int \frac{11}{u}\, du = 11 \int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $11 \log{\left(u \right)}$
      
      The result is: $u + 11 \log{\left(u \right)}$
    Now substitute $u$ back in:
    
    $u + 11 \log{\left(u - 9 \right)} - 9$
  Now substitute $u$ back in:
  
  $3 x + 11 \log{\left(3 x - 9 \right)} - 9$
Method #2
1. Rewrite the integrand:
  
  $\frac{3 x + 2}{x - 3} = 3 + \frac{11}{x - 3}$
2. Integrate term-by-term:
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 3\, dx = 3 x$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{11}{x - 3}\, dx = 11 \int \frac{1}{x - 3}\, dx$
    1. Let $u = x - 3$ .
      
      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:
      
      $\log{\left(x - 3 \right)}$
    So, the result is: $11 \log{\left(x - 3 \right)}$
  The result is: $3 x + 11 \log{\left(x - 3 \right)}$
Method #3
1. Rewrite the integrand:
  
  $\frac{3 x + 2}{x - 3} = \frac{3 x}{x - 3} + \frac{2}{x - 3}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{3 x}{x - 3}\, dx = 3 \int \frac{x}{x - 3}\, dx$
    1. Rewrite the integrand:
      
      $\frac{x}{x - 3} = 1 + \frac{3}{x - 3}$
    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:
      
      $\int \frac{3}{x - 3}\, dx = 3 \int \frac{1}{x - 3}\, dx$
      
      Let $u = x - 3$ .
      
      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:
      
      $\log{\left(x - 3 \right)}$
      
      So, the result is: $3 \log{\left(x - 3 \right)}$
      
      The result is: $x + 3 \log{\left(x - 3 \right)}$
    So, the result is: $3 x + 9 \log{\left(x - 3 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{2}{x - 3}\, dx = 2 \int \frac{1}{x - 3}\, dx$
    1. Let $u = x - 3$ .
      
      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:
      
      $\log{\left(x - 3 \right)}$
    So, the result is: $2 \log{\left(x - 3 \right)}$
  The result is: $3 x + 9 \log{\left(x - 3 \right)} + 2 \log{\left(x - 3 \right)}$
Add the constant of integration:

$3 x + 11 \log{\left(3 x - 9 \right)} - 9+ \mathrm{constant}$

The answer is:

$3 x + 11 \log{\left(3 x - 9 \right)} - 9+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                            
 |                                             
 | 3*x + 2                                     
 | ------- dx = -9 + C + 3*x + 11*log(-9 + 3*x)
 |  x - 3                                      
 |                                             
/

$$\int \frac{3 x + 2}{x - 3}\, dx = C + 3 x + 11 \log{\left(3 x - 9 \right)} - 9$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

3 - 11*log(3) + 11*log(2)

$$- 11 \log{\left(3 \right)} + 3 + 11 \log{\left(2 \right)}$$

3 - 11*log(3) + 11*log(2)

$$- 11 \log{\left(3 \right)} + 3 + 11 \log{\left(2 \right)}$$

3 - 11*log(3) + 11*log(2)

Numerical answer [src]

-1.46011618918981

-1.46011618918981

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 (3x+2)/(x-3) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3