Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of cosx
Integral of x^5/(x^2+1)
Integral of e^(2*x)*cos(3*x)
Integral of 1/(x^2+2x+1)
Identical expressions
(6x- one)/(3x- one)
(6x minus 1) divide by (3x minus 1)
(6x minus one) divide by (3x minus one)
6x-1/3x-1
(6x-1) divide by (3x-1)
(6x-1)/(3x-1)dx
Similar expressions
(6x-1)/(3x+1)
(6x+1)/(3x-1)

Integral of (6x-1)/(3x-1) dx

The solution

You have entered [src]

  0           
  /           
 |            
 |  6*x - 1   
 |  ------- dx
 |  3*x - 1   
 |            
/             
0

$$\int\limits_{0}^{0} \frac{6 x - 1}{3 x - 1}\, dx$$

Integral((6*x - 1)/(3*x - 1), (x, 0, 0))

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

$$0$$

Numerical answer [src]

0.0

0.0

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (6x-1)/(3x-1) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3