# How to calculate $\int \dfrac {e^{t}} {1+t}dt$

The original is slove $\dfrac {df} {dt}+\dfrac {t} {1+t}f=at（ie.a*t）$, IC：$f\left( 0\right) =f_{0}$

I used method of variation of constant and get a indefinite integral $\int \dfrac {et} {1+t}dt$，

it is similar to the indefinite integral $\int \dfrac {e^t} {t}dt$, I use integration by parts and get a infinite series

$e^{t}(t^{-1}-t^{-2}+\ldots +\left( n-1\right) !\left( -1\right) ^{n+1}t^{-n}+\ldots)$,

then I got troubled. I don't know what i should do next.

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is that $e^t$ ?? –  Santosh Linkha Feb 28 '13 at 15:37
Your equation is incomplete.It has nothing on the other end of= Unless that is a*t and not at. –  Ishan Banerjee Feb 28 '13 at 15:37
@experimentX Yes –  Jebei Feb 28 '13 at 15:39
Write $1+t = u$ you get $t = u - 1$ so $e^t = e^u/e$, put the $e$ outside, and you get $\int \frac{e^u}{u} du$, unfortunately the integral is non elementary. Check this out ... however with bounds, you can evaluate it numerically. –  Santosh Linkha Feb 28 '13 at 15:48
talyor expansion? –  user39843 Feb 28 '13 at 15:49

$\int \dfrac{e^t}{1+t}dt=\int \dfrac{e^v}{ev}dv=\dfrac{Ei(v)}{e}+C=\dfrac{Ei(t+1)}{e}+C$ For properties of this function see, http://mathworld.wolfram.com/ExponentialIntegral.html http://functions.wolfram.com/GammaBetaErf/ExpIntegralEi/