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Graphing y =:
(x+1)/(x-1)
Derivative of:
(x+1)/(x-1)
Identical expressions
(x+ one)/(x- one)
(x plus 1) divide by (x minus 1)
(x plus one) divide by (x minus 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+1))
x^2+x+1/(x-1)^3
(x-1)/(x-1)
(-x^2+2x+1)/((x-1)^2*(x^2+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:
    
    $\int \frac{2}{x - 1}\, dx = 2 \int \frac{1}{x - 1}\, dx$
    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: $x + 2 \log{\left(x - 1 \right)}$
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. 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:
      
      $\log{\left(x - 1 \right)}$
    The result is: $x + \log{\left(x - 1 \right)}$
  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: $x + \log{\left(x - 1 \right)} + \log{\left(x - 1 \right)}$
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]

-oo - 2*pi*I

$$-\infty - 2 i \pi$$

-oo - 2*pi*I

$$-\infty - 2 i \pi$$

-oo - 2*pi*i

Numerical answer [src]

-87.181913572439

-87.181913572439

The graph

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2