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Graphing y =:
(1-2x)/(x+1)
Identical expressions
(one -2x)/(x+ one)
(1 minus 2x) divide by (x plus 1)
(one minus 2x) divide by (x plus one)
1-2x/x+1
(1-2x) divide by (x+1)
(1-2x)/(x+1)dx
Similar expressions
(1-2x)/(x-1)
(1+2x)/(x+1)
(1-2*x)/((x+1)^2+4)^(0,5)

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

The solution

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

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

-2 + 3*log(2)

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

-2 + 3*log(2)

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

-2 + 3*log(2)

Numerical answer [src]

0.0794415416798359

0.0794415416798359

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4