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I have this integral:

$$\int e^{-\sqrt{x}}dx.$$

This is what I have done:

$$\int e^{-\sqrt{x}}dx = \int \frac{1}{e^{\sqrt{x}}} dx$$

I Tried to solve it by substitution:

$$t = \sqrt{x}$$ $$ t^2 = x$$ $$ 2d = dx$$

So:

$$\int \frac{1}{e^{\sqrt{x}}} dx = \int \frac{1}{e^t} 2t$$

Then:

$$2\int \frac{t}{e^t}dt$$

How should I proceed?

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  • $\begingroup$ It is $2t \ dt=dx$ not $2 \ d=dx$ $\endgroup$ – callculus Jun 22 '15 at 8:15
  • $\begingroup$ @G.Sassatelli Your edit maybe hide, that the OP has some fundamental problems with substitution. $\endgroup$ – callculus Jun 22 '15 at 8:23
  • $\begingroup$ @calculus It seemed to me that he got the integrand right ($t$ was there), but ok, I'll roll back. $\endgroup$ – user228113 Jun 22 '15 at 8:28
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    $\begingroup$ As an aside, $~\displaystyle\int_0^\infty e^{^{\Large-\sqrt[n]x}}~dx ~=~ n!$ $\endgroup$ – Lucian Jun 22 '15 at 8:32
  • $\begingroup$ If you differentialte both sides of $t^2=x$ you get $2t \ dt=1 \ dx$ Then $\int \frac{1}{e^{\sqrt{x}}} dx$ becomes $\int \frac{1}{e^{t}} 2t \ dt$ $\endgroup$ – callculus Jun 22 '15 at 8:36
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Writing $2\int e^{-t}tdt$, we can use the integration by parts formula $\int fg'=fg-\int f'g$ to get \begin{equation*} -2e^{-t}t+2\int e^{-t}dt. \end{equation*} Use the substitution $u=-t$ to get \begin{equation*} -2e^{-t}t-2\int e^udu. \end{equation*} Integrate and substitute back.

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See the integrand as $te^{-t}$ and integrate by parts.

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  • $\begingroup$ Do you mean that he must see it as $t\,d\left[-e^{-t}\right]$ ? $\endgroup$ – user228113 Jun 22 '15 at 8:32
  • $\begingroup$ Yes. Then integrating by parts, it is just an exponential to integrate. $\endgroup$ – Sylvain L. Jun 22 '15 at 9:15
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Hint: It will be more useful to apply the subsitution $t=-\sqrt{x}$. Then apply integration by parts.

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By parts, $$\int e^{-\sqrt x}dx=\int2\sqrt xe^{-\sqrt x}\frac{dx}{2\sqrt x}=-2\sqrt x e^{-\sqrt x}+\int2e^{-\sqrt x}\frac{dx}{2\sqrt x}=-2\sqrt x e^{-\sqrt x}-2e^{-\sqrt x}.$$

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