Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of x^(1/3)
Integral of sin(x)^4
Integral of cos(x)^3
Integral of (e^(x^2))
Graphing y =:
(2x+3)/(3x-1)
Identical expressions
(2x+ three)/(3x- one)
(2x plus 3) divide by (3x minus 1)
(2x plus three) divide by (3x minus one)
2x+3/3x-1
(2x+3) divide by (3x-1)
(2x+3)/(3x-1)dx
Similar expressions
(2x+3)/(3x+1)
(2x-3)/(3x-1)

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

The solution

You have entered [src]

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

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

nan

$$\text{NaN}$$

nan

$$\text{NaN}$$

nan

Numerical answer [src]

144.399866640766

144.399866640766

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

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3