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Derivative of:
(x^2-9)/(x+3)
How do you in partial fractions?:
(x^2-9)/(x+3)
Graphing y =:
(x^2-9)/(x+3)
Identical expressions
(x^ two - nine)/(x+ three)
(x squared minus 9) divide by (x plus 3)
(x to the power of two minus nine) divide by (x plus three)
(x2-9)/(x+3)
x2-9/x+3
(x²-9)/(x+3)
(x to the power of 2-9)/(x+3)
x^2-9/x+3
(x^2-9) divide by (x+3)
(x^2-9)/(x+3)dx
Similar expressions
(x^2-9)/(x-3)
(x^2+9)/(x+3)

Solving integrals/ (x^2-9)/(x+3)

Integral of (x^2-9)/(x+3) dx

The solution

You have entered [src]

  1          
  /          
 |           
 |   2       
 |  x  - 9   
 |  ------ dx
 |  x + 3    
 |           
/            
0

$$\int\limits_{0}^{1} \frac{x^{2} - 9}{x + 3}\, dx$$

Integral((x^2 - 9)/(x + 3), (x, 0, 1))

Detail solution

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

$\frac{x \left(x - 6\right)}{2}$
Add the constant of integration:

$\frac{x \left(x - 6\right)}{2}+ \mathrm{constant}$

The answer is:

$\frac{x \left(x - 6\right)}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                        
 |                         
 |  2               2      
 | x  - 9          x       
 | ------ dx = C + -- - 3*x
 | x + 3           2       
 |                         
/

$$\int \frac{x^{2} - 9}{x + 3}\, dx = C + \frac{x^{2}}{2} - 3 x$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

-5/2

$$- \frac{5}{2}$$

-5/2

$$- \frac{5}{2}$$

-5/2

Numerical answer [src]

-2.5

-2.5

The graph

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 (x^2-9)/(x+3) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2