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Integral of d{x}:
Integral of k
Integral of -y
Integral of x*e^(2*x)*dx
Integral of (1-x^2)^(1/2)
Identical expressions
(e^x-x)/((e^x*x))
(e to the power of x minus x) divide by ((e to the power of x multiply by x))
(ex-x)/((ex*x))
ex-x/ex*x
(e^x-x)/((e^xx))
(ex-x)/((exx))
ex-x/exx
e^x-x/e^xx
(e^x-x) divide by ((e^x*x))
(e^x-x)/((e^x*x))dx
Similar expressions
(e^x+x)/((e^x*x))
What you mean?
(e^x - x)/((e^x*x))
(e^x - x)/(e^(x*x))
(e^x - x)/(e^(x*x))

You entered:

(e^x-x)/((e^x*x))

What you mean?

(e^x - x)/((e^x*x))

(e^x - x)/(e^(x*x))

(e^x - x)/(e^(x*x))

Integral of (e^x-x)/((e^x*x)) dx

The solution

You have entered [src]

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

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

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

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

$$\log x+e^ {- x }$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

   -1    -2         
- e   + e   + log(2)

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

   -1    -2         
- e   + e   + log(2)

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

Simplify

Numerical answer [src]

0.460603022625116

0.460603022625116

The graph

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

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Identical expressions

Similar expressions

What you mean?

You entered:

What you mean?

Integral of (e^x-x)/((e^x*x)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2