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

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

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

Integral((x + 2)/(x + 1), (x, 0, 1))

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

1 + log(2)

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

=

1 + log(2)

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

1 + log(2)

Numerical answer [src]

1.69314718055995

1.69314718055995

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2