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Identical expressions
(dx)/(e^(x/ two)+e^x)
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(dx) divide by (e to the power of (x divide by two) plus e to the power of x)
(dx)/(e(x/2)+ex)
dx/ex/2+ex
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Similar expressions
(dx)/(e^(x/2)-e^x)

Solving integrals/ (dx)/(e^(x/2)+e^x)

Integral of (dx)/(e^(x/2)+e^x) dx

The solution

You have entered [src]

  1             
  /             
 |              
 |       1      
 |  1*------- dx
 |     x        
 |     -        
 |     2    x   
 |    e  + e    
 |              
/               
0

$$\int\limits_{0}^{1} 1 \cdot \frac{1}{e^{x} + e^{\frac{x}{2}}}\, dx$$

Integral(1/(E^(x/2) + E^x), (x, 0, 1))

Detail solution

Let $u = e^{\frac{x}{2}}$ .

Then let $du = \frac{e^{\frac{x}{2}} dx}{2}$ and substitute $2 du$ :

$\int \frac{4}{u^{3} + u^{2}}\, du$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{2}{u^{3} + u^{2}}\, du = 2 \int \frac{1}{u^{3} + u^{2}}\, du$
  1. Rewrite the integrand:
    
    $\frac{1}{u^{3} + u^{2}} = \frac{1}{u + 1} - \frac{1}{u} + \frac{1}{u^{2}}$
  2. Integrate term-by-term:
    1. Let $u = u + 1$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      Now substitute $u$ back in:
      
      $\log{\left(u + 1 \right)}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u}\right)\, du = - \int \frac{1}{u}\, du$
      1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      So, the result is: $- \log{\left(u \right)}$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$
    The result is: $- \log{\left(u \right)} + \log{\left(u + 1 \right)} - \frac{1}{u}$
  So, the result is: $- 2 \log{\left(u \right)} + 2 \log{\left(u + 1 \right)} - \frac{2}{u}$
Now substitute $u$ back in:

$2 \log{\left(e^{\frac{x}{2}} + 1 \right)} - 2 \log{\left(e^{\frac{x}{2}} \right)} - 2 e^{- \frac{x}{2}}$
Now simplify:

$2 \log{\left(e^{\frac{x}{2}} + 1 \right)} - 2 \log{\left(e^{\frac{x}{2}} \right)} - 2 e^{- \frac{x}{2}}$
Add the constant of integration:

$2 \log{\left(e^{\frac{x}{2}} + 1 \right)} - 2 \log{\left(e^{\frac{x}{2}} \right)} - 2 e^{- \frac{x}{2}}+ \mathrm{constant}$

The answer is:

$2 \log{\left(e^{\frac{x}{2}} + 1 \right)} - 2 \log{\left(e^{\frac{x}{2}} \right)} - 2 e^{- \frac{x}{2}}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

$$2\,\log \left(e^{{{x}\over{2}}}+1\right)-2\,e^ {- {{x}\over{2}} }-x$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

       -1/2                   /     1/2\
1 - 2*e     - 2*log(2) + 2*log\1 + e   /

$$2\,\log \left(\sqrt{e}+1\right)-2\,\log 2-{{2}\over{\sqrt{e}}}+1$$

       -1/2                   /     1/2\
1 - 2*e     - 2*log(2) + 2*log\1 + e   /

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

Упростить

Numerical answer [src]

0.348798287815056

0.348798287815056

The graph

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

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Identical expressions

Similar expressions

Integral of (dx)/(e^(x/2)+e^x) dx

Limits of integration:

The graph:

Piecewise:

The solution