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Identical expressions
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Similar expressions
(y-1)/(y-1)
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Solving integrals/ (y-1)/(y+1)

Integral of (y-1)/(y+1) dx

The solution

You have entered [src]

  1         
  /         
 |          
 |  y - 1   
 |  ----- dy
 |  y + 1   
 |          
/           
0

$$\int\limits_{0}^{1} \frac{y - 1}{y + 1}\, dy$$

Integral((y - 1)/(y + 1), (y, 0, 1))

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                               
 |                                
 | y - 1                          
 | ----- dy = C + y - 2*log(1 + y)
 | y + 1                          
 |                                
/

$$\int \frac{y - 1}{y + 1}\, dy = C + y - 2 \log{\left(y + 1 \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

1 - 2*log(2)

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

1 - 2*log(2)

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

1 - 2*log(2)

Numerical answer [src]

-0.386294361119891

-0.386294361119891

The graph

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (y-1)/(y+1) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2