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Derivative of:
(x^3+1)/(x+2)
How do you in partial fractions?:
(x^3+1)/(x+2)
Identical expressions
(x^ three + one)/(x+ two)
(x cubed plus 1) divide by (x plus 2)
(x to the power of three plus one) divide by (x plus two)
(x3+1)/(x+2)
x3+1/x+2
(x³+1)/(x+2)
(x to the power of 3+1)/(x+2)
x^3+1/x+2
(x^3+1) divide by (x+2)
(x^3+1)/(x+2)dx
Similar expressions
(x^3+1)/(x-2)
(x^3-1)/(x+2)

Integral of (x^3+1)/(x+2) dx

The solution

You have entered [src]

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

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

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{x^{3} + 1}{x + 2} = x^{2} - 2 x + 4 - \frac{7}{x + 2}$
2. Integrate term-by-term:
  1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int x^{2}\, dx = \frac{x^{3}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 x\right)\, dx = - 2 \int x\, dx$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
    So, the result is: $- x^{2}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 4\, dx = 4 x$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{7}{x + 2}\right)\, dx = - 7 \int \frac{1}{x + 2}\, dx$
    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: $- 7 \log{\left(x + 2 \right)}$
  The result is: $\frac{x^{3}}{3} - x^{2} + 4 x - 7 \log{\left(x + 2 \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{x^{3} + 1}{x + 2} = \frac{x^{3}}{x + 2} + \frac{1}{x + 2}$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\frac{x^{3}}{x + 2} = x^{2} - 2 x + 4 - \frac{8}{x + 2}$
  2. Integrate term-by-term:
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{2}\, dx = \frac{x^{3}}{3}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 x\right)\, dx = - 2 \int x\, dx$
      
      The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
      
      So, the result is: $- x^{2}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int 4\, dx = 4 x$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{8}{x + 2}\right)\, dx = - 8 \int \frac{1}{x + 2}\, dx$
      
      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: $- 8 \log{\left(x + 2 \right)}$
    The result is: $\frac{x^{3}}{3} - x^{2} + 4 x - 8 \log{\left(x + 2 \right)}$
  1. Let $u = x + 2$ .
    
    Then let $du = dx$ and substitute $du$ :
    
    $\int \frac{1}{u}\, du$
    1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    Now substitute $u$ back in:
    
    $\log{\left(x + 2 \right)}$
  The result is: $\frac{x^{3}}{3} - x^{2} + 4 x + \log{\left(x + 2 \right)} - 8 \log{\left(x + 2 \right)}$
Add the constant of integration:

$\frac{x^{3}}{3} - x^{2} + 4 x - 7 \log{\left(x + 2 \right)}+ \mathrm{constant}$

The answer is:

$\frac{x^{3}}{3} - x^{2} + 4 x - 7 \log{\left(x + 2 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                            
 |                                             
 |  3                                         3
 | x  + 1           2                        x 
 | ------ dx = C - x  - 7*log(2 + x) + 4*x + --
 | x + 2                                     3 
 |                                             
/

$$\int \frac{x^{3} + 1}{x + 2}\, dx = C + \frac{x^{3}}{3} - x^{2} + 4 x - 7 \log{\left(x + 2 \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

10/3 - 7*log(3) + 7*log(2)

$$- 7 \log{\left(3 \right)} + \frac{10}{3} + 7 \log{\left(2 \right)}$$

10/3 - 7*log(3) + 7*log(2)

$$- 7 \log{\left(3 \right)} + \frac{10}{3} + 7 \log{\left(2 \right)}$$

10/3 - 7*log(3) + 7*log(2)

Numerical answer [src]

0.495077576576183

0.495077576576183

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

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Identical expressions

Similar expressions

Integral of (x^3+1)/(x+2) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2