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dx/(x(x-1)^3)

Integral of dx/(x(x+1)^3) dx

The solution

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  2              
  /              
 |               
 |      1        
 |  ---------- dx
 |           3   
 |  x*(x + 1)    
 |               
/                
1

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

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

Detail solution

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

$\frac{x}{2 \left(x + 1\right)^{3}} + \log{\left(x \right)} - \log{\left(x + 1 \right)} + \frac{1}{x + 1} + \frac{1}{2 \left(x + 1\right)^{3}}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

  17                    
- -- - log(3) + 2*log(2)
  72

$$- \log{\left(3 \right)} - \frac{17}{72} + 2 \log{\left(2 \right)}$$

  17                    
- -- - log(3) + 2*log(2)
  72

$$- \log{\left(3 \right)} - \frac{17}{72} + 2 \log{\left(2 \right)}$$

-17/72 - log(3) + 2*log(2)

Numerical answer [src]

0.0515709613406698

0.0515709613406698

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

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How to use it?

Identical expressions

Similar expressions

Integral of dx/(x(x+1)^3) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3