Question

which is the integral of ∫e^{t²-t}dt? I don't know how to solve the exponential part to equalize with the differential and finally match with the formula
∫e^{u}du=e^{u}+k... i'll appreciate your help

Answer 1

that isn't an elementary integral

Answer 2

please note that:
\[{t^2} - t = {t^2} - t + \frac{1}{4} - \frac{1}{4}\]

Answer 3

Sorry but I don't know how to apply in the operation

Answer 4

\[\int\limits_{}^{}e^{t^2-t} dt \]
this is what you really have?

Answer 5

Yes, that the question

Answer 6

well you could apply what @Michele_Laino said to make it look like that one famous non-elementaty integral
\[\int\limits_{}^{}e^{u^2} du\]
that is
\[t^2-t=t^2-t+\frac{1}{4}-\frac{1}{4}=(t-\frac{1}{2})^2-\frac{1}{4} \\ \int\limits e^{(t-\frac{1}{2})^2-\frac{1}{4}} dt =e^{-\frac{1}{4}} \int\limits e^{(t-\frac{1}{2})^2} dt \]

Answer 7

but there is no elementary way to evaluate that

Answer 8

may be the Gamma or Euler function?

Answer 9

please have you got limits of integration too? @charlymix

Answer 10

NO, actually the question is ∫4t-2(e^{t^{2-t}})dt for a linear equation

Answer 11

oh that is actually an an elementary integral

Answer 12

let u=t^2-t
then du=(2t-1) dt
multiply both sides of the du=(2t-1)=dt by 2
and you hve 2 du =(4t-2) dt