Whats the integral of $e^{1/x}$ dx? Iam trying to solve this as a part of some problem. But I am not able to find the value for this.
Can anyone integrate the above problem.
Please let me know. Thank you.
 A: $$\int e^{1/x} \, dx=e^{1/x} x-\text{Ei}\left(\frac{1}{x}\right) + c$$
where $\operatorname{Ei}(x)$ is the exponential integral function defined as
$$\operatorname{Ei}(x) = - \int_{-x}^\infty \frac{e^{-t} dt}t$$
$\operatorname{Ei}(x)$ is a non-elementary function, and we can only evaluate it using numerical methods.
A: Make a substitution of $x = \tan \theta$, then you obtain
$$ I = \int e^{\cot \theta} \sec^2\theta \ \ d\theta $$
Integrate by parts with $u$ being the exponential:
$$ uv - \int{v \ du} = \tan \theta \ e^{\cot \theta} + \int{\dfrac{e^{\cot \theta}}{\sin \theta \cos \theta} \ d \theta} $$
Make a substitution of $u = \cot \theta \implies du = -\csc^2 \theta$, noting that $\tan \theta = 1/u$ (and substituting $x$ back in):
$$I = xe^{1/x} - \int{\dfrac{e^u}{u} \ du}$$
We arrive at the final answer by substituting everything back in:
$$\boxed{\displaystyle I = xe^{1/x} - \operatorname{Ei} \left( \dfrac{1}{x} \right) + C}$$
I explain more of the small algebraic details here, if any step is unclear.

A note on the solution: There's actually a much simpler solution I found after realizing the trig sub is redundant. Here it is:
Start by integrating by parts, with $dv = 1$:
$$\displaystyle I = xe^{1/x} - \int{\dfrac{-xe^{1/x}}{x^2} \ dx} = xe^{1/x} + \int{ \dfrac{e^{1/x}}{x}  \ dx}$$
We can substitute $u = 1/x \implies du = -1/x^2 \ dx$, noting that $x = 1/u$:
$$\displaystyle I = xe^{1/x} - \int{\dfrac{ue^{u}}{u^2} \ du} = xe^{1/x} - \int{\dfrac{e^{u}}{u} \ du}$$
$$\boxed{\displaystyle I = xe^{1/x} - \operatorname{Ei} \left( \dfrac{1}{x} \right) + C}$$
A: One simple way to do it (if an approximate solution will do) is the following: 
You know the Taylor series of $e^x$ is given by $e^x = 1 + x + \frac{x^2}{2!}+...$. 
Therefore the Taylor series of $e^{1/x}$ is given by $e^{1/x} = 1 + \frac{1}{x} + \frac{1}{2!x^2} +...$
Now just integrate term-by term.
A: There is no definite formula for this you can use series expansion of $e^x$ to compute a numerical value. 
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+....$$
$$\int e^{1/x}=\int1+\frac{1}{x}+\frac{1}{x^22!}+\frac{1}{x^33!}+\frac{1}{x^44!}+.... \\
\int e^x= C + x + \ln x-\frac{1}{2!}\frac{1}{x}-\frac{1}{3!}\frac{1}{2x^2}-\frac{1}{3!}\frac{1}{3x^3}-...$$
