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shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
First take 5t = p 5dt = dp
 2 years ago

gmer Group TitleBest ResponseYou've already chosen the best response.0
\[{1 \over 5} \int {te^p \over (p 1)^2}dp\]
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
you get \[\large \frac{1}{5^2}\int\limits_{}^{}\frac{pe^p}{(p+1)^2}dp\] \[\large \frac{1}{2*5^2}\int\limits_{}^{}\frac{(2p+2)e^p}{(p^2+1+2p)}dp  \large \frac{1}{5^2}\int\limits_{}^{}\frac{e^p}{(p+1)^2}dp\]
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.1
@gmer, got it ??
 2 years ago
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