Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of sin²xcos²x
Integral of tg2x
Integral of 1/(x^2-9)
Integral of x*e^(x*y)
Identical expressions
(exp(2x))/((exp(2x)+ three)^(one / three))
( exponent of (2x)) divide by (( exponent of (2x) plus 3) to the power of (1 divide by 3))
( exponent of (2x)) divide by (( exponent of (2x) plus three) to the power of (one divide by three))
(exp(2x))/((exp(2x)+3)(1/3))
exp2x/exp2x+31/3
exp2x/exp2x+3^1/3
(exp(2x)) divide by ((exp(2x)+3)^(1 divide by 3))
(exp(2x))/((exp(2x)+3)^(1/3))dx
Similar expressions
(exp(2x))/((exp(2x)-3)^(1/3))

Solving integrals/ (exp(2x))/((exp(2x)+3)^(1/3))

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

The solution

You have entered [src]

  1                 
  /                 
 |                  
 |        2*x       
 |       e          
 |  ------------- dx
 |     __________   
 |  3 /  2*x        
 |  \/  e    + 3    
 |                  
/                   
0

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

Integral(exp(2*x)/(exp(2*x) + 3)^(1/3), (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 $\frac{du}{2}$ :
  
  $\int \frac{1}{2 \sqrt[3]{u + 3}}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{\sqrt[3]{u + 3}}\, du = \frac{\int \frac{1}{\sqrt[3]{u + 3}}\, du}{2}$
    1. Let $u = u + 3$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{\sqrt[3]{u}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{\sqrt[3]{u}}\, du = \frac{3 u^{\frac{2}{3}}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{3 \left(u + 3\right)^{\frac{2}{3}}}{2}$
    So, the result is: $\frac{3 \left(u + 3\right)^{\frac{2}{3}}}{4}$
  Now substitute $u$ back in:
  
  $\frac{3 \left(e^{2 x} + 3\right)^{\frac{2}{3}}}{4}$
Method #2
1. Let $u = 2 x$ .
  
  Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
  
  $\int \frac{e^{u}}{2 \sqrt[3]{e^{u} + 3}}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{e^{u}}{\sqrt[3]{e^{u} + 3}}\, du = \frac{\int \frac{e^{u}}{\sqrt[3]{e^{u} + 3}}\, du}{2}$
    1. Let $u = e^{u} + 3$ .
      
      Then let $du = e^{u} du$ and substitute $du$ :
      
      $\int \frac{1}{\sqrt[3]{u}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{\sqrt[3]{u}}\, du = \frac{3 u^{\frac{2}{3}}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{3 \left(e^{u} + 3\right)^{\frac{2}{3}}}{2}$
    So, the result is: $\frac{3 \left(e^{u} + 3\right)^{\frac{2}{3}}}{4}$
  Now substitute $u$ back in:
  
  $\frac{3 \left(e^{2 x} + 3\right)^{\frac{2}{3}}}{4}$
Method #3
1. Let $u = \sqrt[3]{e^{2 x} + 3}$ .
  
  Then let $du = \frac{2 e^{2 x} dx}{3 \left(e^{2 x} + 3\right)^{\frac{2}{3}}}$ and substitute $\frac{3 du}{2}$ :
  
  $\int \frac{3 u}{2}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int u\, du = \frac{3 \int u\, du}{2}$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
    So, the result is: $\frac{3 u^{2}}{4}$
  Now substitute $u$ back in:
  
  $\frac{3 \left(e^{2 x} + 3\right)^{\frac{2}{3}}}{4}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

  /                                      
 |                                    2/3
 |       2*x                /     2*x\   
 |      e                 3*\3 + e   /   
 | ------------- dx = C + ---------------
 |    __________                 4       
 | 3 /  2*x                              
 | \/  e    + 3                          
 |                                       
/

$$\int \frac{e^{2 x}}{\sqrt[3]{e^{2 x} + 3}}\, dx = C + \frac{3 \left(e^{2 x} + 3\right)^{\frac{2}{3}}}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                      2/3
    3 ___     /     2\   
  3*\/ 2    3*\3 + e /   
- ------- + -------------
     2            4

$$- \frac{3 \sqrt[3]{2}}{2} + \frac{3 \left(3 + e^{2}\right)^{\frac{2}{3}}}{4}$$

                      2/3
    3 ___     /     2\   
  3*\/ 2    3*\3 + e /   
- ------- + -------------
     2            4

$$- \frac{3 \sqrt[3]{2}}{2} + \frac{3 \left(3 + e^{2}\right)^{\frac{2}{3}}}{4}$$

-3*2^(1/3)/2 + 3*(3 + exp(2))^(2/3)/4

Numerical answer [src]

1.68102639528624

1.68102639528624

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

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3