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Integral of d{x}:
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Integral of x^2/(1+x)
Derivative of:
(1+x)/(2-x)
Limit of the function:
(1+x)/(2-x)
Identical expressions
(one +x)/(two -x)
(1 plus x) divide by (2 minus x)
(one plus x) divide by (two minus x)
1+x/2-x
(1+x) divide by (2-x)
(1+x)/(2-x)dx
Similar expressions
(1+x)/(2+x)
(1-x)/(2-x)

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

The solution

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

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

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

Detail solution

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

$- x - 3 \log{\left(x - 2 \right)} - 1+ \mathrm{constant}$

The answer is:

$- x - 3 \log{\left(x - 2 \right)} - 1+ \mathrm{constant}$

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

-1 + 3*log(2)

$$-1 + 3 \log{\left(2 \right)}$$

-1 + 3*log(2)

$$-1 + 3 \log{\left(2 \right)}$$

-1 + 3*log(2)

Numerical answer [src]

1.07944154167984

1.07944154167984

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

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4

Method #1

Method #2

Method #3