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Integral of d{x}:
Integral of -e^x
Integral of xe^(1-x)
Integral of x^4*e^(2*x)*dx
Integral of (x²-1)/(x²+1)
Graphing y =:
(2*x+1)/(x+2)
Derivative of:
(2*x+1)/(x+2)
Identical expressions
(two *x+ one)/(x+ two)
(2 multiply by x plus 1) divide by (x plus 2)
(two multiply by x plus one) divide by (x plus two)
(2x+1)/(x+2)
2x+1/x+2
(2*x+1) divide by (x+2)
(2*x+1)/(x+2)dx
Similar expressions
(x^3+2x+1)/(x+2)
(2*x-1)/(x+2)
(2*x+1)/(x-2)
(2x+1)/((x+2)*(x-1))

Solving integrals/ (2*x+1)/(x+2)

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

The solution

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

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

2 - 3*log(3) + 3*log(2)

$$- 3 \log{\left(3 \right)} + 2 + 3 \log{\left(2 \right)}$$

2 - 3*log(3) + 3*log(2)

$$- 3 \log{\left(3 \right)} + 2 + 3 \log{\left(2 \right)}$$

2 - 3*log(3) + 3*log(2)

Numerical answer [src]

0.783604675675507

0.783604675675507

The graph

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3