Mister Exam

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Integral of d{x}:
Integral of √xdx
Integral of √(1-x^2)
Integral of 1/(2x)
Integral of (e^x)/x
Identical expressions
three *e^x*dx/(two *e^x- nine)
3 multiply by e to the power of x multiply by dx divide by (2 multiply by e to the power of x minus 9)
three multiply by e to the power of x multiply by dx divide by (two multiply by e to the power of x minus nine)
3*ex*dx/(2*ex-9)
3*ex*dx/2*ex-9
3e^xdx/(2e^x-9)
3exdx/(2ex-9)
3exdx/2ex-9
3e^xdx/2e^x-9
3*e^x*dx divide by (2*e^x-9)
Similar expressions
3*e^x*dx/(2*e^x+9)

Integral of 3e^xdx/(2*e^x-9) dx

The solution

You have entered [src]

 log(11)           
 -------           
    2              
    /              
   |               
   |         x     
   |      3*E      
   |    -------- dx
   |       x       
   |    2*E  - 9   
   |               
  /                
log(5)

$$\int\limits_{\log{\left(5 \right)}}^{\frac{\log{\left(11 \right)}}{2}} \frac{3 e^{x}}{2 e^{x} - 9}\, dx$$

Integral((3*E^x)/(2*E^x - 9), (x, log(5), log(11)/2))

Detail solution

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

$\frac{3 \log{\left(2 e^{x} - 9 \right)}}{2}+ \mathrm{constant}$

The answer is:

$\frac{3 \log{\left(2 e^{x} - 9 \right)}}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                  
 |                                   
 |      x                 /        x\
 |   3*E             3*log\-9 + 2*e /
 | -------- dx = C + ----------------
 |    x                     2        
 | 2*E  - 9                          
 |                                   
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

nan

$$\text{NaN}$$

nan

$$\text{NaN}$$

nan

Numerical answer [src]

5.48057533727924

5.48057533727924

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

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Identical expressions

Similar expressions

Integral of 3e^xdx/(2*e^x-9) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Integral of 3e^xdx/(2*e^x-9) dx