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Similar expressions
e^x*(1/x-ln(x))

Integral of e^x*(1/x+ln(x)) dx

The solution

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  1                   
  /                   
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 |   x /1         \   
 |  E *|- + log(x)| dx
 |     \x         /   
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/                     
0

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

Integral(E^x*(1/x + log(x)), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = \frac{1}{x}$ .
  
  Then let $du = - \frac{dx}{x^{2}}$ and substitute $- du$ :
  
  $\int \left(- \frac{u e^{\frac{1}{u}} + e^{\frac{1}{u}} \log{\left(\frac{1}{u} \right)}}{u^{2}}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{u e^{\frac{1}{u}} + e^{\frac{1}{u}} \log{\left(\frac{1}{u} \right)}}{u^{2}}\, du = - \int \frac{u e^{\frac{1}{u}} + e^{\frac{1}{u}} \log{\left(\frac{1}{u} \right)}}{u^{2}}\, du$
    1. Let $u = \frac{1}{u}$ .
      
      Then let $du = - \frac{du}{u^{2}}$ and substitute $- du$ :
      
      $\int \left(- \frac{u e^{u} \log{\left(u \right)} + e^{u}}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u e^{u} \log{\left(u \right)} + e^{u}}{u}\, du = - \int \frac{u e^{u} \log{\left(u \right)} + e^{u}}{u}\, du$
      
      Let $u = \log{\left(u \right)}$ .
      
      Then let $du = \frac{du}{u}$ and substitute $du$ :
      
      $\int \left(u e^{u} e^{e^{u}} + e^{e^{u}}\right)\, du$
      
      Integrate term-by-term:
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{u} e^{e^{u}}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = e^{u}$ .
      
      Then let $du = e^{u} du$ and substitute $du$ :
      
      $\int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      Now substitute $u$ back in:
      
      $e^{e^{u}}$
      
      Now evaluate the sub-integral.
      
      Let $u = e^{u}$ .
      
      Then let $du = e^{u} du$ and substitute $du$ :
      
      $\int \frac{e^{u}}{u}\, du$
      
      EiRule(a=1, b=0, context=exp(_u)/_u, symbol=_u)
      
      Now substitute $u$ back in:
      
      $\operatorname{Ei}{\left(e^{u} \right)}$
      
      Let $u = e^{u}$ .
      
      Then let $du = e^{u} du$ and substitute $du$ :
      
      $\int \frac{e^{u}}{u}\, du$
      
      EiRule(a=1, b=0, context=exp(_u)/_u, symbol=_u)
      
      Now substitute $u$ back in:
      
      $\operatorname{Ei}{\left(e^{u} \right)}$
      
      The result is: $u e^{e^{u}}$
      
      Now substitute $u$ back in:
      
      $e^{u} \log{\left(u \right)}$
      
      So, the result is: $- e^{u} \log{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $e^{\frac{1}{u}} \log{\left(u \right)}$
    So, the result is: $- e^{\frac{1}{u}} \log{\left(u \right)}$
  Now substitute $u$ back in:
  
  $e^{x} \log{\left(x \right)}$
Method #2
1. Rewrite the integrand:
  
  $e^{x} \left(\log{\left(x \right)} + \frac{1}{x}\right) = \frac{x e^{x} \log{\left(x \right)} + e^{x}}{x}$
2. Let $u = \frac{1}{x}$ .
  
  Then let $du = - \frac{dx}{x^{2}}$ and substitute $- du$ :
  
  $\int \left(- \frac{u e^{\frac{1}{u}} + e^{\frac{1}{u}} \log{\left(\frac{1}{u} \right)}}{u^{2}}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{u e^{\frac{1}{u}} + e^{\frac{1}{u}} \log{\left(\frac{1}{u} \right)}}{u^{2}}\, du = - \int \frac{u e^{\frac{1}{u}} + e^{\frac{1}{u}} \log{\left(\frac{1}{u} \right)}}{u^{2}}\, du$
    1. Let $u = e^{\frac{1}{u}} \log{\left(\frac{1}{u} \right)}$ .
      
      Then let $du = \left(- \frac{e^{\frac{1}{u}}}{u} - \frac{e^{\frac{1}{u}} \log{\left(\frac{1}{u} \right)}}{u^{2}}\right) du$ and substitute $- du$ :
      
      $\int \left(-1\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      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:
      
      $- e^{\frac{1}{u}} \log{\left(\frac{1}{u} \right)}$
    So, the result is: $e^{\frac{1}{u}} \log{\left(\frac{1}{u} \right)}$
  Now substitute $u$ back in:
  
  $e^{x} \log{\left(x \right)}$
Method #3
1. Rewrite the integrand:
  
  $e^{x} \left(\log{\left(x \right)} + \frac{1}{x}\right) = e^{x} \log{\left(x \right)} + \frac{e^{x}}{x}$
2. Integrate term-by-term:
  1. Let $u = \log{\left(x \right)}$ .
    
    Then let $du = \frac{dx}{x}$ and substitute $du$ :
    
    $\int u e^{u} e^{e^{u}}\, du$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{u} e^{e^{u}}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = e^{u}$ .
      
      Then let $du = e^{u} du$ and substitute $du$ :
      
      $\int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      Now substitute $u$ back in:
      
      $e^{e^{u}}$
      
      Now evaluate the sub-integral.
    2. Let $u = e^{u}$ .
      
      Then let $du = e^{u} du$ and substitute $du$ :
      
      $\int \frac{e^{u}}{u}\, du$
      
      EiRule(a=1, b=0, context=exp(_u)/_u, symbol=_u)
      
      Now substitute $u$ back in:
      
      $\operatorname{Ei}{\left(e^{u} \right)}$
    Now substitute $u$ back in:
    
    $e^{x} \log{\left(x \right)} - \operatorname{Ei}{\left(x \right)}$
  The result is: $e^{x} \log{\left(x \right)}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The answer [src]

oo

$$\infty$$

oo

$$\infty$$

oo

Numerical answer [src]

44.0904461339929

44.0904461339929

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

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How to use it?

Identical expressions

Similar expressions

Integral of e^x*(1/x+ln(x)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3