How to solve $\int e^{-\sqrt{x}}dx$

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

Hint: It will be more useful to apply the subsitution $t=-\sqrt{x}$. Then apply integration by parts.

Answer 4

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