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(two *x+ three)/(two *x+ one)
(2 multiply by x plus 3) divide by (2 multiply by x plus 1)
(two multiply by x plus three) divide by (two multiply by x plus one)
(2x+3)/(2x+1)
2x+3/2x+1
(2*x+3) divide by (2*x+1)
(2*x+3)/(2*x+1)dx
Similar expressions
2x+3/2x+1
(2*x+3)/(2*x-1)
(2x+3/2x+1)
(2*x-3)/(2*x+1)

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

The solution

You have entered [src]

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

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

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

$x + \log{\left(2 x + 1 \right)}+ \mathrm{constant}$

The answer is:

$x + \log{\left(2 x + 1 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

1 + log(3)

$$1 + \log{\left(3 \right)}$$

1 + log(3)

$$1 + \log{\left(3 \right)}$$

1 + log(3)

Numerical answer [src]

2.09861228866811

2.09861228866811

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)/(2x+1) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

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