Question

anti deriv of e^rootx / x

Answer 1

What I would do here, \(\tiny \\[0.9em]\)
\(\color{#000000}{\displaystyle\int\limits_{~}^{~} \frac{e^\sqrt{x~}}{x}~dx=\int\limits_{~}^{~} \frac{e^\sqrt{x}}{\sqrt{x}\cdot \sqrt{x}}~dx=2\int\limits_{~}^{~} \frac{e^\sqrt{x}}{\sqrt{x}}\left(\frac{1}{2\sqrt{x}}dx\right)}\)

Substitution\( : \)
\(\color{#000000}{\displaystyle p=\sqrt{x}}\) and \(\color{#000000}{\displaystyle dp=dx/(2\sqrt{x})}\)

then, your integral becomes, \(\tiny \\[0.9em]\)
\(\color{#000000}{\displaystyle 2\int\limits_{~}^{~} \frac{e^{\color{blue}{\sqrt{x}}}}{\color{blue}{\sqrt{x}}}\left(\color{red}{\frac{1}{2\sqrt{x}}dx}\right)\quad \longrightarrow \quad 2\int\limits_{~}^{~} \frac{e^{\color{blue}{p}}}{\color{blue}{p}}\color{red}{dp}}\)

and then use the fact that,
\(\color{#000000}{\displaystyle e^{{\tiny~}\large p} =\sum_{k=0}^\infty \frac{p^k}{k!}}\)

\(\color{#000000}{\displaystyle 2p^{-1}e^{{\tiny~}\large p} =2\left[\sum_{k=0}^\infty \frac{p^{k-1}}{k!}\right] }\)

\(\color{#000000}{\displaystyle \int 2p^{-1}e^{{\tiny~}\large p}~dp = \int2\left[\sum_{k=0}^\infty \frac{p^{k-1}}{k!}\right]dp=2\left[\sum_{k=0}^\infty \left(\int\frac{p^{k-1}}{k!}~dp\right)\right] }\)

then, of course sub the p back into the result, and +C.

Answer 2

If you have not learned the power-series technique, then I don't really know another way for this indefinite integral.

Answer 3

I've never learned the power series technique?

Answer 4

Are you asking me, lol?
I don't know if you ever learned it before

Answer 5

have you?

Answer 6

Do you still need help with this question?