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

Integral of (4x^2-9)/(2x+3) dx

The solution

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  0            
  /            
 |             
 |     2       
 |  4*x  - 9   
 |  -------- dx
 |  2*x + 3    
 |             
/              
0

$$\int\limits_{0}^{0} \frac{4 x^{2} - 9}{2 x + 3}\, dx$$

Integral((4*x^2 - 9)/(2*x + 3), (x, 0, 0))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{4 x^{2} - 9}{2 x + 3} = 2 x - 3$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 x\, dx = 2 \int x\, dx$
    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:
    
    $\int \left(-3\right)\, dx = - 3 x$
  The result is: $x^{2} - 3 x$
Method #2
1. Rewrite the integrand:
  
  $\frac{4 x^{2} - 9}{2 x + 3} = \frac{4 x^{2}}{2 x + 3} - \frac{9}{2 x + 3}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{4 x^{2}}{2 x + 3}\, dx = 4 \int \frac{x^{2}}{2 x + 3}\, dx$
    1. Rewrite the integrand:
      
      $\frac{x^{2}}{2 x + 3} = \frac{x}{2} - \frac{3}{4} + \frac{9}{4 \left(2 x + 3\right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{x}{2}\, dx = \frac{\int x\, dx}{2}$
      
      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: $\frac{x^{2}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \left(- \frac{3}{4}\right)\, dx = - \frac{3 x}{4}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{9}{4 \left(2 x + 3\right)}\, dx = \frac{9 \int \frac{1}{2 x + 3}\, dx}{4}$
      
      Let $u = 2 x + 3$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{2 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(2 x + 3 \right)}}{2}$
      
      So, the result is: $\frac{9 \log{\left(2 x + 3 \right)}}{8}$
      
      The result is: $\frac{x^{2}}{4} - \frac{3 x}{4} + \frac{9 \log{\left(2 x + 3 \right)}}{8}$
    So, the result is: $x^{2} - 3 x + \frac{9 \log{\left(2 x + 3 \right)}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{9}{2 x + 3}\right)\, dx = - 9 \int \frac{1}{2 x + 3}\, dx$
    1. Let $u = 2 x + 3$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{2 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(2 x + 3 \right)}}{2}$
    So, the result is: $- \frac{9 \log{\left(2 x + 3 \right)}}{2}$
  The result is: $x^{2} - 3 x + \frac{9 \log{\left(2 x + 3 \right)}}{2} - \frac{9 \log{\left(2 x + 3 \right)}}{2}$
Now simplify:

$x \left(x - 3\right)$
Add the constant of integration:

$x \left(x - 3\right)+ \mathrm{constant}$

The answer is:

$x \left(x - 3\right)+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                          
 |                           
 |    2                      
 | 4*x  - 9           2      
 | -------- dx = C + x  - 3*x
 | 2*x + 3                   
 |                           
/

$$\int \frac{4 x^{2} - 9}{2 x + 3}\, dx = C + x^{2} - 3 x$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

$$0$$

Numerical answer [src]

0.0

0.0

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

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Identical expressions

Similar expressions

Integral of (4x^2-9)/(2x+3) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2