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Identical expressions
(one - three x)/(2x+3)
(1 minus 3x) divide by (2x plus 3)
(one minus three x) divide by (2x plus 3)
1-3x/2x+3
(1-3x) divide by (2x+3)
(1-3x)/(2x+3)dx
Similar expressions
(1-3x)/(2x-3)
(1+3x)/(2x+3)

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

The solution

You have entered [src]

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

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

Integral((1 - 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 $du$ :
  
  $\int \frac{u + 1}{2 u - 9}\, du$
  1. Rewrite the integrand:
    
    $\frac{u + 1}{2 u - 9} = \frac{1}{2} + \frac{11}{2 \left(2 u - 9\right)}$
  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:
      
      $\int \frac{11}{2 \left(2 u - 9\right)}\, du = \frac{11 \int \frac{1}{2 u - 9}\, du}{2}$
      
      Let $u = 2 u - 9$ .
      
      Then let $du = 2 du$ 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:
      
      $\frac{\log{\left(2 u - 9 \right)}}{2}$
      
      So, the result is: $\frac{11 \log{\left(2 u - 9 \right)}}{4}$
    The result is: $\frac{u}{2} + \frac{11 \log{\left(2 u - 9 \right)}}{4}$
  Now substitute $u$ back in:
  
  $- \frac{3 x}{2} + \frac{11 \log{\left(- 6 x - 9 \right)}}{4}$
Method #2
1. Rewrite the integrand:
  
  $\frac{1 - 3 x}{2 x + 3} = - \frac{3}{2} + \frac{11}{2 \left(2 x + 3\right)}$
2. Integrate term-by-term:
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \left(- \frac{3}{2}\right)\, dx = - \frac{3 x}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{11}{2 \left(2 x + 3\right)}\, dx = \frac{11 \int \frac{1}{2 x + 3}\, dx}{2}$
    1. Let $u = 2 x + 3$ .
      
      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:
      
      $\frac{\log{\left(2 x + 3 \right)}}{2}$
    So, the result is: $\frac{11 \log{\left(2 x + 3 \right)}}{4}$
  The result is: $- \frac{3 x}{2} + \frac{11 \log{\left(2 x + 3 \right)}}{4}$
Method #3
1. Rewrite the integrand:
  
  $\frac{1 - 3 x}{2 x + 3} = - \frac{3 x - 1}{2 x + 3}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{3 x - 1}{2 x + 3}\right)\, dx = - \int \frac{3 x - 1}{2 x + 3}\, dx$
  1. Let $u = 3 x$ .
    
    Then let $du = 3 dx$ and substitute $du$ :
    
    $\int \frac{u - 1}{2 u + 9}\, du$
    1. Rewrite the integrand:
      
      $\frac{u - 1}{2 u + 9} = \frac{1}{2} - \frac{11}{2 \left(2 u + 9\right)}$
    2. Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{11}{2 \left(2 u + 9\right)}\right)\, du = - \frac{11 \int \frac{1}{2 u + 9}\, du}{2}$
      
      Let $u = 2 u + 9$ .
      
      Then let $du = 2 du$ 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:
      
      $\frac{\log{\left(2 u + 9 \right)}}{2}$
      
      So, the result is: $- \frac{11 \log{\left(2 u + 9 \right)}}{4}$
      
      The result is: $\frac{u}{2} - \frac{11 \log{\left(2 u + 9 \right)}}{4}$
    Now substitute $u$ back in:
    
    $\frac{3 x}{2} - \frac{11 \log{\left(6 x + 9 \right)}}{4}$
  So, the result is: $- \frac{3 x}{2} + \frac{11 \log{\left(6 x + 9 \right)}}{4}$
Method #4
1. Rewrite the integrand:
  
  $\frac{1 - 3 x}{2 x + 3} = - \frac{3 x}{2 x + 3} + \frac{1}{2 x + 3}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{3 x}{2 x + 3}\right)\, dx = - 3 \int \frac{x}{2 x + 3}\, dx$
    1. Rewrite the integrand:
      
      $\frac{x}{2 x + 3} = \frac{1}{2} - \frac{3}{2 \left(2 x + 3\right)}$
    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:
      
      $\int \left(- \frac{3}{2 \left(2 x + 3\right)}\right)\, dx = - \frac{3 \int \frac{1}{2 x + 3}\, dx}{2}$
      
      Let $u = 2 x + 3$ .
      
      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:
      
      $\frac{\log{\left(2 x + 3 \right)}}{2}$
      
      So, the result is: $- \frac{3 \log{\left(2 x + 3 \right)}}{4}$
      
      The result is: $\frac{x}{2} - \frac{3 \log{\left(2 x + 3 \right)}}{4}$
    So, the result is: $- \frac{3 x}{2} + \frac{9 \log{\left(2 x + 3 \right)}}{4}$
  1. Let $u = 2 x + 3$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{1}{2 u}\, du$
    1. 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:
    
    $\frac{\log{\left(2 x + 3 \right)}}{2}$
  The result is: $- \frac{3 x}{2} + \frac{\log{\left(2 x + 3 \right)}}{2} + \frac{9 \log{\left(2 x + 3 \right)}}{4}$
Add the constant of integration:

$- \frac{3 x}{2} + \frac{11 \log{\left(- 6 x - 9 \right)}}{4}+ \mathrm{constant}$

The answer is:

$- \frac{3 x}{2} + \frac{11 \log{\left(- 6 x - 9 \right)}}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                       
 |                                        
 | 1 - 3*x          3*x   11*log(-9 - 6*x)
 | ------- dx = C - --- + ----------------
 | 2*x + 3           2           4        
 |                                        
/

$$\int \frac{1 - 3 x}{2 x + 3}\, dx = C - \frac{3 x}{2} + \frac{11 \log{\left(- 6 x - 9 \right)}}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  3   11*log(3)   11*log(5)
- - - --------- + ---------
  2       4           4

$$- \frac{11 \log{\left(3 \right)}}{4} - \frac{3}{2} + \frac{11 \log{\left(5 \right)}}{4}$$

  3   11*log(3)   11*log(5)
- - - --------- + ---------
  2       4           4

$$- \frac{11 \log{\left(3 \right)}}{4} - \frac{3}{2} + \frac{11 \log{\left(5 \right)}}{4}$$

-3/2 - 11*log(3)/4 + 11*log(5)/4

Numerical answer [src]

-0.0952295346435256

-0.0952295346435256

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4