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Graphing y =:
(x^2-1)/(x-1)
How do you in partial fractions?:
(x^2-1)/(x-1)
Identical expressions
(x^ two - one)/(x- one)
(x squared minus 1) divide by (x minus 1)
(x to the power of two minus one) divide by (x minus one)
(x2-1)/(x-1)
x2-1/x-1
(x²-1)/(x-1)
(x to the power of 2-1)/(x-1)
x^2-1/x-1
(x^2-1) divide by (x-1)
(x^2-1)/(x-1)dx
Similar expressions
(x^2+1)/(x-1)
(x^2-1)/(x+1)

Solving integrals/ (x^2-1)/(x-1)

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

The solution

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

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

Integral((x^2 - 1*1)/(x - 1*1), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{x^{2} - 1}{x - 1} = x + 1$
2. Integrate term-by-term:
  1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int x\, dx = \frac{x^{2}}{2}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 1\, dx = x$
  The result is: $\frac{x^{2}}{2} + x$
Method #2
1. Rewrite the integrand:
  
  $\frac{x^{2} - 1}{x - 1} = \frac{x^{2}}{x - 1} - \frac{1}{x - 1}$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\frac{x^{2}}{x - 1} = x + 1 + \frac{1}{x - 1}$
  2. Integrate term-by-term:
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
    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: $\frac{x^{2}}{2} + x + \log{\left(x - 1 \right)}$
  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: $\frac{x^{2}}{2} + x$
Now simplify:

$\frac{x \left(x + 2\right)}{2}$
Add the constant of integration:

$\frac{x \left(x + 2\right)}{2}+ \mathrm{constant}$

The answer is:

$\frac{x \left(x + 2\right)}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                      
 |                       
 |  2                   2
 | x  - 1              x 
 | ------ dx = C + x + --
 | x - 1               2 
 |                       
/

$${{x^2+2\,x}\over{2}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

3/2

$${{3}\over{2}}$$

3/2

$$\frac{3}{2}$$

Simplify

Numerical answer [src]

1.5

1.5

The graph

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2