How to solve integral f(u) = e^(u/u-1)*[1/(1-u)^2] ?
I will retype this clear in my reply. Thanks

Question

anonymous · Answer

$$\int\limits_{-∞}^{+∞} e^{\frac{ u }{ u-1 }} * \frac{ 1 }{ (1-u)^2 }$$

dumbcow · Answer

hmm lets try a substitution...
$$w = \frac{u}{u-1}$$
$$dw = \frac{-1}{(u-1)^{2}} du$$
note:
$$(u-1)^{2} = (1-u)^{2}$$
$$\rightarrow - \int\limits e^{w} dw$$

anonymous · Answer

so, which means it equal to $$-e^{w} + C$$ ?

dumbcow · Answer

yes...sub back in to get in terms of "u"

anonymous · Answer

Thanks @dumbcow may I ask in further  if 0<u<1, means w also 0<w<1
so is that means $$-\int\limits_{0}^{1} e^{w} = 1?$$

dumbcow · Answer

no the limits change whenever you make a substitution....thats why after integrating you sub back in the original variable

you can also change the limits but i prefer not to

zzr0ck3r · Answer

same here

dumbcow · Answer

$$-\int\limits e^{w} = -e^{w} = -e^{u/(u-1)}|_0^1$$

anonymous · Answer

so $$-e^{1/0} + e^{0} = \infty + 1 ?$$

dumbcow · Answer

$$-\infty +1$$
:)

anonymous · Answer

Thanks :)