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

The solution

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

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

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{x}{x + 1} \left(x - 2\right)^{2} = x^{2} - 5 x + 9 - \frac{9}{x + 1}$
2. Integrate term-by-term:
  1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int x^{2}\, dx = \frac{x^{3}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 5 x\right)\, dx = - 5 \int x\, dx$
    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: $- \frac{5 x^{2}}{2}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 9\, dx = 9 x$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{9}{x + 1}\right)\, dx = - 9 \int \frac{1}{x + 1}\, dx$
    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: $- 9 \log{\left(x + 1 \right)}$
  The result is: $\frac{x^{3}}{3} - \frac{5 x^{2}}{2} + 9 x - 9 \log{\left(x + 1 \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{x}{x + 1} \left(x - 2\right)^{2} = \frac{x^{3} - 4 x^{2} + 4 x}{x + 1}$
2. Rewrite the integrand:
  
  $\frac{x^{3} - 4 x^{2} + 4 x}{x + 1} = x^{2} - 5 x + 9 - \frac{9}{x + 1}$
3. Integrate term-by-term:
  1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int x^{2}\, dx = \frac{x^{3}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 5 x\right)\, dx = - 5 \int x\, dx$
    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: $- \frac{5 x^{2}}{2}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 9\, dx = 9 x$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{9}{x + 1}\right)\, dx = - 9 \int \frac{1}{x + 1}\, dx$
    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: $- 9 \log{\left(x + 1 \right)}$
  The result is: $\frac{x^{3}}{3} - \frac{5 x^{2}}{2} + 9 x - 9 \log{\left(x + 1 \right)}$
Method #3
1. Rewrite the integrand:
  
  $\frac{x}{x + 1} \left(x - 2\right)^{2} = \frac{x^{3}}{x + 1} - \frac{4 x^{2}}{x + 1} + \frac{4 x}{x + 1}$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\frac{x^{3}}{x + 1} = x^{2} - x + 1 - \frac{1}{x + 1}$
  2. Integrate term-by-term:
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{2}\, dx = \frac{x^{3}}{3}$
    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:
      
      $\int 1\, dx = x$
    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: $\frac{x^{3}}{3} - \frac{x^{2}}{2} + x - \log{\left(x + 1 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{4 x^{2}}{x + 1}\right)\, dx = - 4 \int \frac{x^{2}}{x + 1}\, dx$
    1. Rewrite the integrand:
      
      $\frac{x^{2}}{x + 1} = x - 1 + \frac{1}{x + 1}$
    2. Integrate term-by-term:
      
      The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \left(-1\right)\, dx = - x$
      
      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: $\frac{x^{2}}{2} - x + \log{\left(x + 1 \right)}$
    So, the result is: $- 2 x^{2} + 4 x - 4 \log{\left(x + 1 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{4 x}{x + 1}\, dx = 4 \int \frac{x}{x + 1}\, dx$
    1. Rewrite the integrand:
      
      $\frac{x}{x + 1} = 1 - \frac{1}{x + 1}$
    2. Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
      
      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: $4 x - 4 \log{\left(x + 1 \right)}$
  The result is: $\frac{x^{3}}{3} - \frac{5 x^{2}}{2} + 9 x - 9 \log{\left(x + 1 \right)}$
Add the constant of integration:

$\frac{x^{3}}{3} - \frac{5 x^{2}}{2} + 9 x - 9 \log{\left(x + 1 \right)}+ \mathrm{constant}$

The answer is:

$\frac{x^{3}}{3} - \frac{5 x^{2}}{2} + 9 x - 9 \log{\left(x + 1 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                      
 |                                                 2    3
 |   x          2                               5*x    x 
 | -----*(x - 2)  dx = C - 9*log(1 + x) + 9*x - ---- + --
 | x + 1                                         2     3 
 |                                                       
/

$$\int \frac{x}{x + 1} \left(x - 2\right)^{2}\, dx = C + \frac{x^{3}}{3} - \frac{5 x^{2}}{2} + 9 x - 9 \log{\left(x + 1 \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

41/6 - 9*log(2)

$$\frac{41}{6} - 9 \log{\left(2 \right)}$$

41/6 - 9*log(2)

$$\frac{41}{6} - 9 \log{\left(2 \right)}$$

41/6 - 9*log(2)

Numerical answer [src]

0.595008708293826

0.595008708293826

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

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3