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Integral of (x^2-81)/(x+9)
Integral of tg^4(3x)
Identical expressions
(x^ two - eighty-one)/(x+ nine)
(x squared minus 81) divide by (x plus 9)
(x to the power of two minus eighty minus one) divide by (x plus nine)
(x2-81)/(x+9)
x2-81/x+9
(x²-81)/(x+9)
(x to the power of 2-81)/(x+9)
x^2-81/x+9
(x^2-81) divide by (x+9)
(x^2-81)/(x+9)dx
Similar expressions
(x^2+81)/(x+9)
(x^2-81)/(x-9)

Solving integrals/ (x^2-81)/(x+9)

Integral of (x^2-81)/(x+9) dx

The solution

You have entered [src]

  1           
  /           
 |            
 |   2        
 |  x  - 81   
 |  ------- dx
 |   x + 9    
 |            
/             
0

$$\int\limits_{0}^{1} \frac{x^{2} - 81}{x + 9}\, dx$$

Integral((x^2 - 81)/(x + 9), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{x^{2} - 81}{x + 9} = x - 9$
2. Integrate term-by-term:
  1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int x\, dx = \frac{x^{2}}{2}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \left(-9\right)\, dx = - 9 x$
  The result is: $\frac{x^{2}}{2} - 9 x$
Method #2
1. Rewrite the integrand:
  
  $\frac{x^{2} - 81}{x + 9} = \frac{x^{2}}{x + 9} - \frac{81}{x + 9}$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\frac{x^{2}}{x + 9} = x - 9 + \frac{81}{x + 9}$
  2. Integrate term-by-term:
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \left(-9\right)\, dx = - 9 x$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{81}{x + 9}\, dx = 81 \int \frac{1}{x + 9}\, dx$
      
      Let $u = x + 9$ .
      
      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:
      
      $\log{\left(x + 9 \right)}$
      
      So, the result is: $81 \log{\left(x + 9 \right)}$
    The result is: $\frac{x^{2}}{2} - 9 x + 81 \log{\left(x + 9 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{81}{x + 9}\right)\, dx = - 81 \int \frac{1}{x + 9}\, dx$
    1. Let $u = x + 9$ .
      
      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:
      
      $\log{\left(x + 9 \right)}$
    So, the result is: $- 81 \log{\left(x + 9 \right)}$
  The result is: $\frac{x^{2}}{2} - 9 x + 81 \log{\left(x + 9 \right)} - 81 \log{\left(x + 9 \right)}$
Now simplify:

$\frac{x \left(x - 18\right)}{2}$
Add the constant of integration:

$\frac{x \left(x - 18\right)}{2}+ \mathrm{constant}$

The answer is:

$\frac{x \left(x - 18\right)}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                         
 |                          
 |  2                2      
 | x  - 81          x       
 | ------- dx = C + -- - 9*x
 |  x + 9           2       
 |                          
/

$$\int \frac{x^{2} - 81}{x + 9}\, dx = C + \frac{x^{2}}{2} - 9 x$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

-17/2

$$- \frac{17}{2}$$

-17/2

$$- \frac{17}{2}$$

-17/2

Numerical answer [src]

-8.5

-8.5

The graph

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

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Identical expressions

Similar expressions

Integral of (x^2-81)/(x+9) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2