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integral of $e^frac{1}{sqrt{x}}$

Integral of $frac{-2}{sqrt{16-x^2}}$Calculus Question: Improper integral $int_{0}^{infty}frac{cos(2x+1)}{sqrt[3]{x}}dx$Indefinite integral of $frac{sqrt{x}}{sqrt{x}+1}$How to evaluate the integral of $sqrt{sinsqrt x}cos sqrt x / ( 1+x^2)$?Another messy integral: $I=int frac{sqrt{2-x-x^2}}{x^2} dx$could this integral be solved using substitution?Evaluate $int ( 3x + frac{6x^2 sin^2(frac{x}{2})}{x - sin x} ) frac{(x-sin x)^{3/2}}{sqrt{x}} mathrm{d}x$Evaluating the Integral: $int(frac{ln(sqrt{x})}{x}dx$Evaluating $int sqrt{frac{3-x}{3+x}} sin^{-1}left(sqrt{frac{3-x}{6}}right)dx$Evaluate $int frac{ln(t+sqrt{t^2+1)}}{1+t^2} , dt$

0

$begingroup$

Is there a way to evaluate $int e^frac{1}{sqrt{x}}dx$

I first used substitution, I let $frac{1}{sqrt{x}}=uimplies frac{-1}{2}x^frac{-3}{2}dx=du$ and also $x^frac{-3}{2}=u^3$ so $dx=-2u^{-3}du$

After substituting, I now have to evaluate
$-2int u^{-3}e^u du$ .

I then used integration by parts and ended up with
$-2u^{-3}e^u+u^{-1}e^u-2e^uln u+2int e^uln u du$

This is where I got the problem

calculus

share|cite|improve this question

asked Mar 18 at 13:36

mamotebangmamotebang

152

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  $begingroup$
  Is this a part of finding a definite integral, or are you looking for an anti-derivative? I don't think there is an elementary anti-derivative, so it will probably involve special functions.
  $endgroup$
  – StackTD
  Mar 18 at 13:38
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, thank you StackTD
  $endgroup$
  – mamotebang
  Mar 18 at 13:45
  

  
  
  
  


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  $begingroup$
  $displaystyle int e^{x} ln(x) dx = e^{x} ln(x) - int frac{e^x}{x} dx$, and $displaystyleint_{-infty}^x frac{e^t}{t}dt = text{Ei}(x)$ a special function, see fr.wikipedia.org/wiki/Exponentielle_int%C3%A9grale
  $endgroup$
  – LAGRIDA
  Mar 18 at 13:56
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Is there a way to evaluate $int e^frac{1}{sqrt{x}}dx$

I first used substitution, I let $frac{1}{sqrt{x}}=uimplies frac{-1}{2}x^frac{-3}{2}dx=du$ and also $x^frac{-3}{2}=u^3$ so $dx=-2u^{-3}du$

After substituting, I now have to evaluate
$-2int u^{-3}e^u du$ .

I then used integration by parts and ended up with
$-2u^{-3}e^u+u^{-1}e^u-2e^uln u+2int e^uln u du$

This is where I got the problem

calculus

share|cite|improve this question

asked Mar 18 at 13:36

mamotebangmamotebang

152

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this a part of finding a definite integral, or are you looking for an anti-derivative? I don't think there is an elementary anti-derivative, so it will probably involve special functions.
  $endgroup$
  – StackTD
  Mar 18 at 13:38
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, thank you StackTD
  $endgroup$
  – mamotebang
  Mar 18 at 13:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $displaystyle int e^{x} ln(x) dx = e^{x} ln(x) - int frac{e^x}{x} dx$, and $displaystyleint_{-infty}^x frac{e^t}{t}dt = text{Ei}(x)$ a special function, see fr.wikipedia.org/wiki/Exponentielle_int%C3%A9grale
  $endgroup$
  – LAGRIDA
  Mar 18 at 13:56
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Is there a way to evaluate $int e^frac{1}{sqrt{x}}dx$

I first used substitution, I let $frac{1}{sqrt{x}}=uimplies frac{-1}{2}x^frac{-3}{2}dx=du$ and also $x^frac{-3}{2}=u^3$ so $dx=-2u^{-3}du$

After substituting, I now have to evaluate
$-2int u^{-3}e^u du$ .

I then used integration by parts and ended up with
$-2u^{-3}e^u+u^{-1}e^u-2e^uln u+2int e^uln u du$

This is where I got the problem

calculus

share|cite|improve this question

asked Mar 18 at 13:36

mamotebangmamotebang

152

$endgroup$

share|cite|improve this question

asked Mar 18 at 13:36

mamotebangmamotebang

152

share|cite|improve this question

asked Mar 18 at 13:36

mamotebangmamotebang

152

asked Mar 18 at 13:36

mamotebangmamotebang

152

asked Mar 18 at 13:36

mamotebangmamotebang

152