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Identical expressions
(exp(x)- one)/exp(x)+ one
( exponent of (x) minus 1) divide by exponent of (x) plus 1
( exponent of (x) minus one) divide by exponent of (x) plus one
expx-1/expx+1
(exp(x)-1) divide by exp(x)+1
(exp(x)-1)/exp(x)+1dx
Similar expressions
(exp(x)+1)/exp(x)+1
(exp(x)-1)/exp(x)-1

Integral of (exp(x)-1)/exp(x)+1 dx

The solution

You have entered [src]

  1                
  /                
 |                 
 |  / x        \   
 |  |e  - 1    |   
 |  |------ + 1| dx
 |  |   x      |   
 |  \  e       /   
 |                 
/                  
0

$$\int\limits_{0}^{1} \left(\frac{e^{x} - 1}{e^{x}} + 1\right)\, dx$$

Integral((exp(x) - 1)/exp(x) + 1, (x, 0, 1))

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

     -1
1 + e

$$e^{-1} + 1$$

     -1
1 + e

$$e^{-1} + 1$$

1 + exp(-1)

Numerical answer [src]

1.36787944117144

1.36787944117144

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

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Identical expressions

Similar expressions

Integral of (exp(x)-1)/exp(x)+1 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2