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Integral of d{x}:
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Integral of x^2/4
Integral of x^2*e^(x^3)
Integral of x^(-3/4)
Identical expressions
((z^ two)+ one)/(z+ two)(z- six)
((z squared ) plus 1) divide by (z plus 2)(z minus 6)
((z to the power of two) plus one) divide by (z plus two)(z minus six)
((z2)+1)/(z+2)(z-6)
z2+1/z+2z-6
((z²)+1)/(z+2)(z-6)
((z to the power of 2)+1)/(z+2)(z-6)
z^2+1/z+2z-6
((z^2)+1) divide by (z+2)(z-6)
Similar expressions
((z^2)-1)/(z+2)(z-6)
((z^2)+1)/(z+2)(z+6)
((z^2)+1)/(z-2)(z-6)

Solving integrals/ ((z^2)+1)/(z+2)(z-6)

Integral of ((z^2)+1)/(z+2)(z-6) dz

The solution

You have entered [src]

  1                    
  /                    
 |                     
 |  / 2    \           
 |  \z  + 1/*(z - 6)   
 |  ---------------- dz
 |       z + 2         
 |                     
/                      
0

$$\int\limits_{0}^{1} \frac{\left(z - 6\right) \left(z^{2} + 1\right)}{z + 2}\, dz$$

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{\left(z - 6\right) \left(z^{2} + 1\right)}{z + 2} = z^{2} - 8 z + 17 - \frac{40}{z + 2}$
2. Integrate term-by-term:
  1. The integral of $z^{n}$ is $\frac{z^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int z^{2}\, dz = \frac{z^{3}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 8 z\right)\, dz = - 8 \int z\, dz$
    1. The integral of $z^{n}$ is $\frac{z^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int z\, dz = \frac{z^{2}}{2}$
    So, the result is: $- 4 z^{2}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 17\, dz = 17 z$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{40}{z + 2}\right)\, dz = - 40 \int \frac{1}{z + 2}\, dz$
    1. Let $u = z + 2$ .
      
      Then let $du = dz$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(z + 2 \right)}$
    So, the result is: $- 40 \log{\left(z + 2 \right)}$
  The result is: $\frac{z^{3}}{3} - 4 z^{2} + 17 z - 40 \log{\left(z + 2 \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{\left(z - 6\right) \left(z^{2} + 1\right)}{z + 2} = \frac{z^{3} - 6 z^{2} + z - 6}{z + 2}$
2. Rewrite the integrand:
  
  $\frac{z^{3} - 6 z^{2} + z - 6}{z + 2} = z^{2} - 8 z + 17 - \frac{40}{z + 2}$
3. Integrate term-by-term:
  1. The integral of $z^{n}$ is $\frac{z^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int z^{2}\, dz = \frac{z^{3}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 8 z\right)\, dz = - 8 \int z\, dz$
    1. The integral of $z^{n}$ is $\frac{z^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int z\, dz = \frac{z^{2}}{2}$
    So, the result is: $- 4 z^{2}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 17\, dz = 17 z$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{40}{z + 2}\right)\, dz = - 40 \int \frac{1}{z + 2}\, dz$
    1. Let $u = z + 2$ .
      
      Then let $du = dz$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(z + 2 \right)}$
    So, the result is: $- 40 \log{\left(z + 2 \right)}$
  The result is: $\frac{z^{3}}{3} - 4 z^{2} + 17 z - 40 \log{\left(z + 2 \right)}$
Method #3
1. Rewrite the integrand:
  
  $\frac{\left(z - 6\right) \left(z^{2} + 1\right)}{z + 2} = \frac{z^{3}}{z + 2} - \frac{6 z^{2}}{z + 2} + \frac{z}{z + 2} - \frac{6}{z + 2}$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\frac{z^{3}}{z + 2} = z^{2} - 2 z + 4 - \frac{8}{z + 2}$
  2. Integrate term-by-term:
    1. The integral of $z^{n}$ is $\frac{z^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int z^{2}\, dz = \frac{z^{3}}{3}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 z\right)\, dz = - 2 \int z\, dz$
      
      The integral of $z^{n}$ is $\frac{z^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int z\, dz = \frac{z^{2}}{2}$
      
      So, the result is: $- z^{2}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int 4\, dz = 4 z$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{8}{z + 2}\right)\, dz = - 8 \int \frac{1}{z + 2}\, dz$
      
      Let $u = z + 2$ .
      
      Then let $du = dz$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(z + 2 \right)}$
      
      So, the result is: $- 8 \log{\left(z + 2 \right)}$
    The result is: $\frac{z^{3}}{3} - z^{2} + 4 z - 8 \log{\left(z + 2 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{6 z^{2}}{z + 2}\right)\, dz = - 6 \int \frac{z^{2}}{z + 2}\, dz$
    1. Rewrite the integrand:
      
      $\frac{z^{2}}{z + 2} = z - 2 + \frac{4}{z + 2}$
    2. Integrate term-by-term:
      
      The integral of $z^{n}$ is $\frac{z^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int z\, dz = \frac{z^{2}}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \left(-2\right)\, dz = - 2 z$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{4}{z + 2}\, dz = 4 \int \frac{1}{z + 2}\, dz$
      
      Let $u = z + 2$ .
      
      Then let $du = dz$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(z + 2 \right)}$
      
      So, the result is: $4 \log{\left(z + 2 \right)}$
      
      The result is: $\frac{z^{2}}{2} - 2 z + 4 \log{\left(z + 2 \right)}$
    So, the result is: $- 3 z^{2} + 12 z - 24 \log{\left(z + 2 \right)}$
  1. Rewrite the integrand:
    
    $\frac{z}{z + 2} = 1 - \frac{2}{z + 2}$
  2. Integrate term-by-term:
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dz = z$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{2}{z + 2}\right)\, dz = - 2 \int \frac{1}{z + 2}\, dz$
      
      Let $u = z + 2$ .
      
      Then let $du = dz$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(z + 2 \right)}$
      
      So, the result is: $- 2 \log{\left(z + 2 \right)}$
    The result is: $z - 2 \log{\left(z + 2 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{6}{z + 2}\right)\, dz = - 6 \int \frac{1}{z + 2}\, dz$
    1. Let $u = z + 2$ .
      
      Then let $du = dz$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(z + 2 \right)}$
    So, the result is: $- 6 \log{\left(z + 2 \right)}$
  The result is: $\frac{z^{3}}{3} - 4 z^{2} + 17 z - 34 \log{\left(z + 2 \right)} - 6 \log{\left(z + 2 \right)}$
Add the constant of integration:

$\frac{z^{3}}{3} - 4 z^{2} + 17 z - 40 \log{\left(z + 2 \right)}+ \mathrm{constant}$

The answer is:

$\frac{z^{3}}{3} - 4 z^{2} + 17 z - 40 \log{\left(z + 2 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                          
 |                                                           
 | / 2    \                                                 3
 | \z  + 1/*(z - 6)                             2          z 
 | ---------------- dz = C - 40*log(2 + z) - 4*z  + 17*z + --
 |      z + 2                                              3 
 |                                                           
/

$${{z^3-12\,z^2+51\,z}\over{3}}-40\,\log \left(z+2\right)$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

40/3 - 40*log(3) + 40*log(2)

$$40\,\log 2-{{120\,\log 3-40}\over{3}}$$

40/3 - 40*log(3) + 40*log(2)

$$- 40 \log{\left(3 \right)} + \frac{40}{3} + 40 \log{\left(2 \right)}$$

Simplify

Numerical answer [src]

-2.88527099099324

-2.88527099099324

The graph

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

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Identical expressions

Similar expressions

Integral of ((z^2)+1)/(z+2)(z-6) dz

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3