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Integral of d{x}:
Integral of cos^2xsinxdx
Integral of (2x+3)/(4x-9)
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Integral of cos^5xsinx
Identical expressions
(2x+ three)/(4x- nine)
(2x plus 3) divide by (4x minus 9)
(2x plus three) divide by (4x minus nine)
2x+3/4x-9
(2x+3) divide by (4x-9)
(2x+3)/(4x-9)dx
Similar expressions
(2x-3)/(4x-9)
(2x+3)/(4x+9)

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

The solution

You have entered [src]

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

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

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = 2 x$ .
  
  Then let $du = 2 dx$ and substitute $du$ :
  
  $\int \frac{u + 3}{4 u - 18}\, du$
  1. Rewrite the integrand:
    
    $\frac{u + 3}{4 u - 18} = \frac{1}{4} + \frac{15}{4 \left(2 u - 9\right)}$
  2. Integrate term-by-term:
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, du = \frac{u}{4}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{15}{4 \left(2 u - 9\right)}\, du = \frac{15 \int \frac{1}{2 u - 9}\, du}{4}$
      
      Let $u = 2 u - 9$ .
      
      Then let $du = 2 du$ 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:
      
      $\frac{\log{\left(2 u - 9 \right)}}{2}$
      
      So, the result is: $\frac{15 \log{\left(2 u - 9 \right)}}{8}$
    The result is: $\frac{u}{4} + \frac{15 \log{\left(2 u - 9 \right)}}{8}$
  Now substitute $u$ back in:
  
  $\frac{x}{2} + \frac{15 \log{\left(4 x - 9 \right)}}{8}$
Method #2
1. Rewrite the integrand:
  
  $\frac{2 x + 3}{4 x - 9} = \frac{1}{2} + \frac{15}{2 \left(4 x - 9\right)}$
2. Integrate term-by-term:
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{2}\, dx = \frac{x}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{15}{2 \left(4 x - 9\right)}\, dx = \frac{15 \int \frac{1}{4 x - 9}\, dx}{2}$
    1. Let $u = 4 x - 9$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{1}{4 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{4}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(4 x - 9 \right)}}{4}$
    So, the result is: $\frac{15 \log{\left(4 x - 9 \right)}}{8}$
  The result is: $\frac{x}{2} + \frac{15 \log{\left(4 x - 9 \right)}}{8}$
Method #3
1. Rewrite the integrand:
  
  $\frac{2 x + 3}{4 x - 9} = \frac{2 x}{4 x - 9} + \frac{3}{4 x - 9}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{2 x}{4 x - 9}\, dx = 2 \int \frac{x}{4 x - 9}\, dx$
    1. Rewrite the integrand:
      
      $\frac{x}{4 x - 9} = \frac{1}{4} + \frac{9}{4 \left(4 x - 9\right)}$
    2. Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, dx = \frac{x}{4}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{9}{4 \left(4 x - 9\right)}\, dx = \frac{9 \int \frac{1}{4 x - 9}\, dx}{4}$
      
      Let $u = 4 x - 9$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{1}{4 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{4}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(4 x - 9 \right)}}{4}$
      
      So, the result is: $\frac{9 \log{\left(4 x - 9 \right)}}{16}$
      
      The result is: $\frac{x}{4} + \frac{9 \log{\left(4 x - 9 \right)}}{16}$
    So, the result is: $\frac{x}{2} + \frac{9 \log{\left(4 x - 9 \right)}}{8}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{3}{4 x - 9}\, dx = 3 \int \frac{1}{4 x - 9}\, dx$
    1. Let $u = 4 x - 9$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{1}{4 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{4}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(4 x - 9 \right)}}{4}$
    So, the result is: $\frac{3 \log{\left(4 x - 9 \right)}}{4}$
  The result is: $\frac{x}{2} + \frac{3 \log{\left(4 x - 9 \right)}}{4} + \frac{9 \log{\left(4 x - 9 \right)}}{8}$
Add the constant of integration:

$\frac{x}{2} + \frac{15 \log{\left(4 x - 9 \right)}}{8}+ \mathrm{constant}$

The answer is:

$\frac{x}{2} + \frac{15 \log{\left(4 x - 9 \right)}}{8}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                     
 |                                      
 | 2*x + 3          x   15*log(-9 + 4*x)
 | ------- dx = C + - + ----------------
 | 4*x - 9          2          8        
 |                                      
/

$$\int \frac{2 x + 3}{4 x - 9}\, dx = C + \frac{x}{2} + \frac{15 \log{\left(4 x - 9 \right)}}{8}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

     15*pi*I
oo + -------
        8

$$\infty + \frac{15 i \pi}{8}$$

     15*pi*I
oo + -------
        8

$$\infty + \frac{15 i \pi}{8}$$

oo + 15*pi*i/8

Numerical answer [src]

82.2922428334831

82.2922428334831

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3