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

The solution

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  3              
  /              
 |               
 |   2           
 |  x  + x - 2   
 |  ---------- dx
 |    x - 1      
 |               
/                
1

$$\int\limits_{1}^{3} \frac{\left(x^{2} + x\right) - 2}{x - 1}\, dx$$

Integral((x^2 + x - 2)/(x - 1), (x, 1, 3))

Detail solution

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

$\frac{x \left(x + 4\right)}{2}$
Add the constant of integration:

$\frac{x \left(x + 4\right)}{2}+ \mathrm{constant}$

The answer is:

$\frac{x \left(x + 4\right)}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

$$8$$

Numerical answer [src]

8.0

8.0

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

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Identical expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2