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Graphing y =:
(x^2-16)/(x+4)
Identical expressions
(x^ two - sixteen)/(x+ four)
(x squared minus 16) divide by (x plus 4)
(x to the power of two minus sixteen) divide by (x plus four)
(x2-16)/(x+4)
x2-16/x+4
(x²-16)/(x+4)
(x to the power of 2-16)/(x+4)
x^2-16/x+4
(x^2-16) divide by (x+4)
(x^2-16)/(x+4)dx
Similar expressions
(x^2+16)/(x+4)
(x^2-16)/(x-4)

Solving integrals/ (x^2-16)/(x+4)

Integral of (x^2-16)/(x+4) dx

The solution

You have entered [src]

  1           
  /           
 |            
 |   2        
 |  x  - 16   
 |  ------- dx
 |   x + 4    
 |            
/             
0

$$\int\limits_{0}^{1} \frac{x^{2} - 16}{x + 4}\, dx$$

Integral((x^2 - 16)/(x + 4), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{x^{2} - 16}{x + 4} = x - 4$
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:
    
    $\int \left(-4\right)\, dx = - 4 x$
  The result is: $\frac{x^{2}}{2} - 4 x$
Method #2
1. Rewrite the integrand:
  
  $\frac{x^{2} - 16}{x + 4} = \frac{x^{2}}{x + 4} - \frac{16}{x + 4}$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\frac{x^{2}}{x + 4} = x - 4 + \frac{16}{x + 4}$
  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:
      
      $\int \left(-4\right)\, dx = - 4 x$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{16}{x + 4}\, dx = 16 \int \frac{1}{x + 4}\, dx$
      
      Let $u = x + 4$ .
      
      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:
      
      $\log{\left(x + 4 \right)}$
      
      So, the result is: $16 \log{\left(x + 4 \right)}$
    The result is: $\frac{x^{2}}{2} - 4 x + 16 \log{\left(x + 4 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{16}{x + 4}\right)\, dx = - 16 \int \frac{1}{x + 4}\, dx$
    1. Let $u = x + 4$ .
      
      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:
      
      $\log{\left(x + 4 \right)}$
    So, the result is: $- 16 \log{\left(x + 4 \right)}$
  The result is: $\frac{x^{2}}{2} - 4 x - 16 \log{\left(x + 4 \right)} + 16 \log{\left(x + 4 \right)}$
Now simplify:

$\frac{x \left(x - 8\right)}{2}$
Add the constant of integration:

$\frac{x \left(x - 8\right)}{2}+ \mathrm{constant}$

The answer is:

$\frac{x \left(x - 8\right)}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                         
 |                          
 |  2                2      
 | x  - 16          x       
 | ------- dx = C + -- - 4*x
 |  x + 4           2       
 |                          
/

$$\int \frac{x^{2} - 16}{x + 4}\, dx = C + \frac{x^{2}}{2} - 4 x$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

-7/2

$$- \frac{7}{2}$$

-7/2

$$- \frac{7}{2}$$

-7/2

Numerical answer [src]

-3.5

-3.5

The graph

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

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Identical expressions

Similar expressions

Integral of (x^2-16)/(x+4) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2