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Identical expressions
(x^ two -x- six)/(one -x)
(x squared minus x minus 6) divide by (1 minus x)
(x to the power of two minus x minus six) divide by (one minus x)
(x2-x-6)/(1-x)
x2-x-6/1-x
(x²-x-6)/(1-x)
(x to the power of 2-x-6)/(1-x)
x^2-x-6/1-x
(x^2-x-6) divide by (1-x)
(x^2-x-6)/(1-x)dx
Similar expressions
(x^2-x+6)/(1-x)
(x^2+x-6)/(1-x)
(x^2-x-6)/(1+x)

Integral of (x^2-x-6)/(1-x) dx

The solution

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  1              
  /              
 |               
 |   2           
 |  x  - x - 6   
 |  ---------- dx
 |    1 - x      
 |               
/                
0

$$\int\limits_{0}^{1} \frac{x^{2} - x - 6}{- x + 1}\, dx$$

Integral((x^2 - x - 1*6)/(1 - x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = - x$ .
  
  Then let $du = - dx$ and substitute $du$ :
  
  $\int \frac{- u^{2} - u + 6}{u + 1}\, du$
  1. Let $u = - u$ .
    
    Then let $du = - du$ and substitute $du$ :
    
    $\int \frac{- u^{2} + u + 6}{u - 1}\, du$
    1. Rewrite the integrand:
      
      $\frac{- u^{2} + u + 6}{u - 1} = - u + \frac{6}{u - 1}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u\right)\, du = - \int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $- \frac{u^{2}}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{6}{u - 1}\, du = 6 \int \frac{1}{u - 1}\, du$
      
      Let $u = u - 1$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(u - 1 \right)}$
      
      So, the result is: $6 \log{\left(u - 1 \right)}$
      
      The result is: $- \frac{u^{2}}{2} + 6 \log{\left(u - 1 \right)}$
    Now substitute $u$ back in:
    
    $- \frac{u^{2}}{2} + 6 \log{\left(- u - 1 \right)}$
  Now substitute $u$ back in:
  
  $- \frac{x^{2}}{2} + 6 \log{\left(x - 1 \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{x^{2} - x - 6}{1 - x} = - x + \frac{6}{x - 1}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- x\right)\, dx = - \int x\, dx$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
    So, the result is: $- \frac{x^{2}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{6}{x - 1}\, dx = 6 \int \frac{1}{x - 1}\, dx$
    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: $6 \log{\left(x - 1 \right)}$
  The result is: $- \frac{x^{2}}{2} + 6 \log{\left(x - 1 \right)}$
Method #3
1. Rewrite the integrand:
  
  $\frac{x^{2} - x - 6}{1 - x} = \frac{- x^{2} + x + 6}{x - 1}$
2. Rewrite the integrand:
  
  $\frac{- x^{2} + x + 6}{x - 1} = - x + \frac{6}{x - 1}$
3. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- x\right)\, dx = - \int x\, dx$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
    So, the result is: $- \frac{x^{2}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{6}{x - 1}\, dx = 6 \int \frac{1}{x - 1}\, dx$
    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: $6 \log{\left(x - 1 \right)}$
  The result is: $- \frac{x^{2}}{2} + 6 \log{\left(x - 1 \right)}$
Method #4
1. Rewrite the integrand:
  
  $\frac{x^{2} - x - 6}{1 - x} = \frac{x^{2}}{1 - x} - \frac{x}{1 - x} - \frac{6}{1 - x}$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\frac{x^{2}}{1 - x} = - x - 1 - \frac{1}{x - 1}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- x\right)\, dx = - \int x\, dx$
      
      The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
      
      So, the result is: $- \frac{x^{2}}{2}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \left(-1\right)\, dx = - x$
    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: $- \frac{x^{2}}{2} - x - \log{\left(x - 1 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{x}{1 - x}\right)\, dx = - \int \frac{x}{1 - x}\, dx$
    1. Rewrite the integrand:
      
      $\frac{x}{1 - x} = -1 - \frac{1}{x - 1}$
    2. Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \left(-1\right)\, dx = - x$
      
      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: $- x - \log{\left(x - 1 \right)}$
    So, 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:
    
    $\int \left(- \frac{6}{1 - x}\right)\, dx = - 6 \int \frac{1}{1 - x}\, dx$
    1. There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = 1 - x$ .
      
      Then let $du = - dx$ and substitute $- du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u}\right)\, du = - \int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $- \log{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \log{\left(1 - x \right)}$
      
      Method #2
      
      Rewrite the integrand:
      
      $\frac{1}{1 - x} = - \frac{1}{x - 1}$
      
      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: $- \log{\left(x - 1 \right)}$
      
      Method #3
      
      Rewrite the integrand:
      
      $\frac{1}{1 - x} = - \frac{1}{x - 1}$
      
      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: $- \log{\left(x - 1 \right)}$
    So, the result is: $6 \log{\left(1 - x \right)}$
  The result is: $- \frac{x^{2}}{2} + 6 \log{\left(1 - x \right)}$
Add the constant of integration:

$- \frac{x^{2}}{2} + 6 \log{\left(x - 1 \right)}+ \mathrm{constant}$

The answer is:

$- \frac{x^{2}}{2} + 6 \log{\left(x - 1 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                      
 |                                       
 |  2                                   2
 | x  - x - 6                          x 
 | ---------- dx = C + 6*log(-1 + x) - --
 |   1 - x                             2 
 |                                       
/

$$6\,\log \left(x-1\right)-{{x^2}\over{2}}$$

Simplify

The answer [src]

-oo - 6*pi*I

$${\it \%a}$$

-oo - 6*pi*I

$$-\infty - 6 i \pi$$

Simplify

Numerical answer [src]

-265.045740717317

-265.045740717317

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

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Identical expressions

Similar expressions

Integral of (x^2-x-6)/(1-x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4

Method #1

Method #2

Method #3