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Integral of dx/(3-2*x)
Identical expressions
dx/(three - two *x)
dx divide by (3 minus 2 multiply by x)
dx divide by (three minus two multiply by x)
dx/(3-2x)
dx/3-2x
dx divide by (3-2*x)
Similar expressions
(dx)/(3-2x^2)^-2
dx/3-2x
dx/(3-2x)^2
dx/(3+2*x)
dx/(3-2x+x^2)^1/2
36dx/(3-2x)^3

Solving integrals/ dx/(3-2*x)

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

The solution

You have entered [src]

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

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

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

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

log(3)
------
  2

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

log(3)
------
  2

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

log(3)/2

Numerical answer [src]

0.549306144334055

0.549306144334055

The graph

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3