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Identical expressions
(x^ two +4x- two)/(x^ three -4x)
(x squared plus 4x minus 2) divide by (x cubed minus 4x)
(x to the power of two plus 4x minus two) divide by (x to the power of three minus 4x)
(x2+4x-2)/(x3-4x)
x2+4x-2/x3-4x
(x²+4x-2)/(x³-4x)
(x to the power of 2+4x-2)/(x to the power of 3-4x)
x^2+4x-2/x^3-4x
(x^2+4x-2) divide by (x^3-4x)
(x^2+4x-2)/(x^3-4x)dx
Similar expressions
(x^2-4x-2)/(x^3-4x)
(x^2+4x-2)/(x^3+4x)
(x^2+4x+2)/(x^3-4x)

Integral of (x^2+4x-2)/(x^3-4x) dx

The solution

You have entered [src]

  1                
  /                
 |                 
 |   2             
 |  x  + 4*x - 2   
 |  ------------ dx
 |     3           
 |    x  - 4*x     
 |                 
/                  
0

$$\int\limits_{0}^{1} \frac{x^{2} + 4 x - 2}{x^{3} - 4 x}\, dx$$

Integral((x^2 + 4*x - 1*2)/(x^3 - 4*x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{x^{2} + 4 x - 2}{x^{3} - 4 x} = - \frac{3}{4 \left(x + 2\right)} + \frac{5}{4 \left(x - 2\right)} + \frac{1}{2 x}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{3}{4 \left(x + 2\right)}\right)\, dx = - \frac{3 \int \frac{1}{x + 2}\, dx}{4}$
    1. Let $u = x + 2$ .
      
      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:
      
      $\log{\left(x + 2 \right)}$
    So, the result is: $- \frac{3 \log{\left(x + 2 \right)}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{5}{4 \left(x - 2\right)}\, dx = \frac{5 \int \frac{1}{x - 2}\, dx}{4}$
    1. Let $u = x - 2$ .
      
      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:
      
      $\log{\left(x - 2 \right)}$
    So, the result is: $\frac{5 \log{\left(x - 2 \right)}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{2 x}\, dx = \frac{\int \frac{1}{x}\, dx}{2}$
    1. The integral of $\frac{1}{x}$ is $\log{\left(x \right)}$ .
    So, the result is: $\frac{\log{\left(x \right)}}{2}$
  The result is: $\frac{\log{\left(x \right)}}{2} + \frac{5 \log{\left(x - 2 \right)}}{4} - \frac{3 \log{\left(x + 2 \right)}}{4}$
Method #2
1. Rewrite the integrand:
  
  $\frac{x^{2} + 4 x - 2}{x^{3} - 4 x} = \frac{x^{2}}{x^{3} - 4 x} + \frac{4 x}{x^{3} - 4 x} - \frac{2}{x^{3} - 4 x}$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\frac{x^{2}}{x^{3} - 4 x} = \frac{1}{2 \left(x + 2\right)} + \frac{1}{2 \left(x - 2\right)}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{2 \left(x + 2\right)}\, dx = \frac{\int \frac{1}{x + 2}\, dx}{2}$
      
      Let $u = x + 2$ .
      
      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:
      
      $\log{\left(x + 2 \right)}$
      
      So, the result is: $\frac{\log{\left(x + 2 \right)}}{2}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{2 \left(x - 2\right)}\, dx = \frac{\int \frac{1}{x - 2}\, dx}{2}$
      
      Let $u = x - 2$ .
      
      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:
      
      $\log{\left(x - 2 \right)}$
      
      So, the result is: $\frac{\log{\left(x - 2 \right)}}{2}$
    The result is: $\frac{\log{\left(x - 2 \right)}}{2} + \frac{\log{\left(x + 2 \right)}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{4 x}{x^{3} - 4 x}\, dx = 4 \int \frac{x}{x^{3} - 4 x}\, dx$
    1. Rewrite the integrand:
      
      $\frac{x}{x^{3} - 4 x} = - \frac{1}{4 \left(x + 2\right)} + \frac{1}{4 \left(x - 2\right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{4 \left(x + 2\right)}\right)\, dx = - \frac{\int \frac{1}{x + 2}\, dx}{4}$
      
      Let $u = x + 2$ .
      
      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:
      
      $\log{\left(x + 2 \right)}$
      
      So, the result is: $- \frac{\log{\left(x + 2 \right)}}{4}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{4 \left(x - 2\right)}\, dx = \frac{\int \frac{1}{x - 2}\, dx}{4}$
      
      Let $u = x - 2$ .
      
      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:
      
      $\log{\left(x - 2 \right)}$
      
      So, the result is: $\frac{\log{\left(x - 2 \right)}}{4}$
      
      The result is: $\frac{\log{\left(x - 2 \right)}}{4} - \frac{\log{\left(x + 2 \right)}}{4}$
    So, the result is: $\log{\left(x - 2 \right)} - \log{\left(x + 2 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{2}{x^{3} - 4 x}\right)\, dx = - 2 \int \frac{1}{x^{3} - 4 x}\, dx$
    1. Rewrite the integrand:
      
      $\frac{1}{x^{3} - 4 x} = \frac{1}{8 \left(x + 2\right)} + \frac{1}{8 \left(x - 2\right)} - \frac{1}{4 x}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{8 \left(x + 2\right)}\, dx = \frac{\int \frac{1}{x + 2}\, dx}{8}$
      
      Let $u = x + 2$ .
      
      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:
      
      $\log{\left(x + 2 \right)}$
      
      So, the result is: $\frac{\log{\left(x + 2 \right)}}{8}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{8 \left(x - 2\right)}\, dx = \frac{\int \frac{1}{x - 2}\, dx}{8}$
      
      Let $u = x - 2$ .
      
      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:
      
      $\log{\left(x - 2 \right)}$
      
      So, the result is: $\frac{\log{\left(x - 2 \right)}}{8}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{4 x}\right)\, dx = - \frac{\int \frac{1}{x}\, dx}{4}$
      
      The integral of $\frac{1}{x}$ is $\log{\left(x \right)}$ .
      
      So, the result is: $- \frac{\log{\left(x \right)}}{4}$
      
      The result is: $- \frac{\log{\left(x \right)}}{4} + \frac{\log{\left(x - 2 \right)}}{8} + \frac{\log{\left(x + 2 \right)}}{8}$
    So, the result is: $\frac{\log{\left(x \right)}}{2} - \frac{\log{\left(x - 2 \right)}}{4} - \frac{\log{\left(x + 2 \right)}}{4}$
  The result is: $\frac{\log{\left(x \right)}}{2} + \frac{5 \log{\left(x - 2 \right)}}{4} - \frac{3 \log{\left(x + 2 \right)}}{4}$
Add the constant of integration:

$\frac{\log{\left(x \right)}}{2} + \frac{5 \log{\left(x - 2 \right)}}{4} - \frac{3 \log{\left(x + 2 \right)}}{4}+ \mathrm{constant}$

The answer is:

$\frac{\log{\left(x \right)}}{2} + \frac{5 \log{\left(x - 2 \right)}}{4} - \frac{3 \log{\left(x + 2 \right)}}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                           
 |                                                            
 |  2                                                         
 | x  + 4*x - 2          log(x)   3*log(2 + x)   5*log(-2 + x)
 | ------------ dx = C + ------ - ------------ + -------------
 |    3                    2           4               4      
 |   x  - 4*x                                                 
 |                                                            
/

$$-{{3\,\log \left(x+2\right)}\over{4}}+{{\log x}\over{2}}+{{5\,\log \left(x-2\right)}\over{4}}$$

Simplify

The answer [src]

     5*pi*I
oo + ------
       4

$${\it \%a}$$

     5*pi*I
oo + ------
       4

$$\infty + \frac{5 i \pi}{4}$$

Simplify

Numerical answer [src]

20.8746902602154

20.8746902602154

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

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Identical expressions

Similar expressions

Integral of (x^2+4x-2)/(x^3-4x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2