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Q.
If $\int_0^1 \frac{ e ^{ t } dt }{1+ t }= A$ then the value of $\int_0^1 \frac{ e ^{ t } dt }{(1+ t )^2}$ is :
Integrals
Solution:
$I=\int\limits_0^1 \frac{e^t(t+1-t)}{(1+t)^2} d t=\int\limits_0^1 \frac{e^t}{1+t} d t-\int\limits_0^1 e^t\left(\frac{1}{1+t}-\frac{1}{(1+t)^2}\right) d t$
$\left.= A -\frac{ e ^{ t }}{1+ t }\right]_0^1= A -\frac{ e }{2}+1$; Alternatively I. B. P. directly.