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

The solution

You have entered [src]

  1                    
  /                    
 |                     
 |         1           
 |  ---------------- dx
 |         2           
 |  (x - 1) *(x - 2)   
 |                     
/                      
0

$$\int\limits_{0}^{1} \frac{1}{\left(x - 2\right) \left(x - 1\right)^{2}}\, dx$$

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{1}{\left(x - 2\right) \left(x - 1\right)^{2}} = - \frac{1}{x - 1} - \frac{1}{\left(x - 1\right)^{2}} + \frac{1}{x - 2}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{x - 1}\right)\, dx = - \int \frac{1}{x - 1}\, dx$
    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: $- \log{\left(x - 1 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{\left(x - 1\right)^{2}}\right)\, dx = - \int \frac{1}{\left(x - 1\right)^{2}}\, dx$
    1. Let $u = x - 1$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u^{2}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$
      
      Now substitute $u$ back in:
      
      $- \frac{1}{x - 1}$
    So, the result is: $\frac{1}{x - 1}$
  1. Let $u = x - 2$ .
    
    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:
    
    $\log{\left(x - 2 \right)}$
  The result is: $\log{\left(x - 2 \right)} - \log{\left(x - 1 \right)} + \frac{1}{x - 1}$
Method #2
1. Rewrite the integrand:
  
  $\frac{1}{\left(x - 2\right) \left(x - 1\right)^{2}} = \frac{1}{x^{3} - 4 x^{2} + 5 x - 2}$
2. Rewrite the integrand:
  
  $\frac{1}{x^{3} - 4 x^{2} + 5 x - 2} = - \frac{1}{x - 1} - \frac{1}{\left(x - 1\right)^{2}} + \frac{1}{x - 2}$
3. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{x - 1}\right)\, dx = - \int \frac{1}{x - 1}\, dx$
    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: $- \log{\left(x - 1 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{\left(x - 1\right)^{2}}\right)\, dx = - \int \frac{1}{\left(x - 1\right)^{2}}\, dx$
    1. Let $u = x - 1$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u^{2}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$
      
      Now substitute $u$ back in:
      
      $- \frac{1}{x - 1}$
    So, the result is: $\frac{1}{x - 1}$
  1. Let $u = x - 2$ .
    
    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:
    
    $\log{\left(x - 2 \right)}$
  The result is: $\log{\left(x - 2 \right)} - \log{\left(x - 1 \right)} + \frac{1}{x - 1}$
Method #3
1. Rewrite the integrand:
  
  $\frac{1}{\left(x - 2\right) \left(x - 1\right)^{2}} = \frac{1}{x^{3} - 4 x^{2} + 5 x - 2}$
2. Rewrite the integrand:
  
  $\frac{1}{x^{3} - 4 x^{2} + 5 x - 2} = - \frac{1}{x - 1} - \frac{1}{\left(x - 1\right)^{2}} + \frac{1}{x - 2}$
3. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{x - 1}\right)\, dx = - \int \frac{1}{x - 1}\, dx$
    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: $- \log{\left(x - 1 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{\left(x - 1\right)^{2}}\right)\, dx = - \int \frac{1}{\left(x - 1\right)^{2}}\, dx$
    1. Let $u = x - 1$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u^{2}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$
      
      Now substitute $u$ back in:
      
      $- \frac{1}{x - 1}$
    So, the result is: $\frac{1}{x - 1}$
  1. Let $u = x - 2$ .
    
    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:
    
    $\log{\left(x - 2 \right)}$
  The result is: $\log{\left(x - 2 \right)} - \log{\left(x - 1 \right)} + \frac{1}{x - 1}$
Now simplify:

$\frac{\left(x - 1\right) \left(\log{\left(x - 2 \right)} - \log{\left(x - 1 \right)}\right) + 1}{x - 1}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The answer [src]

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$$-\infty$$

-oo

$$-\infty$$

-oo

Numerical answer [src]

-1.38019561125665e+19

-1.38019561125665e+19

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

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3