Question

Integrate \[\int{te^{-5t} \over (5t-1)^2} dt\]

Answer 1

First take -5t = p
-5dt = dp

Answer 2

\[-{1 \over 5} \int {te^p \over (-p -1)^2}dp\]

Answer 3

you get
\[\large \frac{1}{5^2}\int\limits_{}^{}\frac{pe^p}{(p+1)^2}dp\]
\[\large \frac{1}{2*5^2}\int\limits_{}^{}\frac{(2p+2)e^p}{(p^2+1+2p)}dp - \large \frac{1}{5^2}\int\limits_{}^{}\frac{e^p}{(p+1)^2}dp\]

Answer 4

@gmer, got it ??