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Integral of e^(2x)/(1-3e^(2x))
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Identical expressions
e^(2x)/(one -3e^(2x))
e to the power of (2x) divide by (1 minus 3e to the power of (2x))
e to the power of (2x) divide by (one minus 3e to the power of (2x))
e(2x)/(1-3e(2x))
e2x/1-3e2x
e^2x/1-3e^2x
e^(2x) divide by (1-3e^(2x))
e^(2x)/(1-3e^(2x))dx
Similar expressions
e^(2x)/(1+3e^(2x))

Integral of e^(2x)/(1-3e^(2x)) dx

The solution

You have entered [src]

  1              
  /              
 |               
 |      2*x      
 |     E         
 |  ---------- dx
 |         2*x   
 |  1 - 3*E      
 |               
/                
0

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

Integral(E^(2*x)/(1 - 3*exp(2*x)), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = e^{2 x}$ .
  
  Then let $du = 2 e^{2 x} dx$ and substitute $- du$ :
  
  $\int \left(- \frac{1}{6 u - 2}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{6 u - 2}\, du = - \int \frac{1}{6 u - 2}\, du$
    1. There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = 6 u - 2$ .
      
      Then let $du = 6 du$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{1}{6 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{6}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(6 u - 2 \right)}}{6}$
      
      Method #2
      
      Rewrite the integrand:
      
      $\frac{1}{6 u - 2} = \frac{1}{2 \left(3 u - 1\right)}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{2 \left(3 u - 1\right)}\, du = \frac{\int \frac{1}{3 u - 1}\, du}{2}$
      
      Let $u = 3 u - 1$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{1}{3 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{3}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(3 u - 1 \right)}}{3}$
      
      So, the result is: $\frac{\log{\left(3 u - 1 \right)}}{6}$
    So, the result is: $- \frac{\log{\left(6 u - 2 \right)}}{6}$
  Now substitute $u$ back in:
  
  $- \frac{\log{\left(6 e^{2 x} - 2 \right)}}{6}$
Method #2
1. Rewrite the integrand:
  
  $\frac{e^{2 x}}{1 - 3 e^{2 x}} = - \frac{e^{2 x}}{3 e^{2 x} - 1}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{e^{2 x}}{3 e^{2 x} - 1}\right)\, dx = - \int \frac{e^{2 x}}{3 e^{2 x} - 1}\, dx$
  1. Let $u = 3 e^{2 x} - 1$ .
    
    Then let $du = 6 e^{2 x} dx$ and substitute $\frac{du}{6}$ :
    
    $\int \frac{1}{6 u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{6}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{6}$
    Now substitute $u$ back in:
    
    $\frac{\log{\left(3 e^{2 x} - 1 \right)}}{6}$
  So, the result is: $- \frac{\log{\left(3 e^{2 x} - 1 \right)}}{6}$
Add the constant of integration:

$- \frac{\log{\left(6 e^{2 x} - 2 \right)}}{6}+ \mathrm{constant}$

The answer is:

$- \frac{\log{\left(6 e^{2 x} - 2 \right)}}{6}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                    
 |                                     
 |     2*x                /        2*x\
 |    E                log\-2 + 6*e   /
 | ---------- dx = C - ----------------
 |        2*x                 6        
 | 1 - 3*E                             
 |                                     
/

$$\int \frac{e^{2 x}}{1 - 3 e^{2 x}}\, dx = C - \frac{\log{\left(6 e^{2 x} - 2 \right)}}{6}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

     /  1    2\           
  log|- - + e |           
     \  3     /   log(2/3)
- ------------- + --------
        6            6

$$- \frac{\log{\left(- \frac{1}{3} + e^{2} \right)}}{6} + \frac{\log{\left(\frac{2}{3} \right)}}{6}$$

     /  1    2\           
  log|- - + e |           
     \  3     /   log(2/3)
- ------------- + --------
        6            6

$$- \frac{\log{\left(- \frac{1}{3} + e^{2} \right)}}{6} + \frac{\log{\left(\frac{2}{3} \right)}}{6}$$

-log(-1/3 + exp(2))/6 + log(2/3)/6

Numerical answer [src]

-0.393217355908468

-0.393217355908468

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

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Identical expressions

Similar expressions

Integral of e^(2x)/(1-3e^(2x)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2