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anonymous
 4 years ago
Integrate \[\int{te^{5t} \over (5t1)^2} dt\]
anonymous
 4 years ago
Integrate \[\int{te^{5t} \over (5t1)^2} dt\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First take 5t = p 5dt = dp

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you get \[\large \frac{1}{5^2}\int\limits_{}^{}\frac{pe^p}{(p+1)^2}dp\] \[\large \frac{1}{2*5^2}\int\limits_{}^{}\frac{(2p+2)e^p}{(p^2+1+2p)}dp  \large \frac{1}{5^2}\int\limits_{}^{}\frac{e^p}{(p+1)^2}dp\]
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