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Integral of 1/(2-x)
Derivative of:
(x-1)/(x+1)
Graphing y =:
(x-1)/(x+1)
Canonical form:
(x-1)/(x+1)
Identical expressions
(x- one)/(x+ one)
(x minus 1) divide by (x plus 1)
(x minus one) divide by (x plus one)
x-1/x+1
(x-1) divide by (x+1)
(x-1)/(x+1)dx
Similar expressions
(x+1)/(x+1)
(x-1)/((x+1)((x^2)-4))
(x-1)/(x-1)
1/x(√x-1/x+1)

Solving integrals/ (x-1)/(x+1)

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

The solution

You have entered [src]

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

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

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

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

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.

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How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2