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gmer
Integrate \[\int{te^{-5t} \over (5t-1)^2} dt\]
First take -5t = p -5dt = dp
\[-{1 \over 5} \int {te^p \over (-p -1)^2}dp\]
you get \[\large \frac{1}{5^2}\int\limits_{}^{}\frac{pe^p}{(p+1)^2}dp\] \[\large \frac{1}{2*5^2}\int\limits_{}^{}\frac{(2p+2)e^p}{(p^2+1+2p)}dp - \large \frac{1}{5^2}\int\limits_{}^{}\frac{e^p}{(p+1)^2}dp\]