8 Answers
kunl
·2011-04-16 21:37:12
i am unable to integrate the second term we on integrating by parts..
we get second term like this
-∫e-x/(1+e2x)
chatur
·2011-04-17 00:19:48
put e^x=t \rightarrow \int \frac{\cot^{-1}t}{t^2} dt\rightarrow cot^{-1}t\left ( \frac{-1}{t} \right )-\int \left ( \frac{-1}{1+t^2} \int \frac{1}{t^2}dt\right )dt
\rightarrow cot^{-1}t \left ( \frac{-1}{t} \right )-\int \left ( \frac{1}{(t)(1+t^2)} \right ) \rightarrow cot^{-1}t \left ( \frac{-1}{t} \right )-\int \left \left ( \frac{1}{t} -\frac{t}{1+t^2}\right )dt
\rightarrow cot^{-1}t \left ( \frac{-1}{t} \right )-lnt+\frac{1}{2}ln(1+t^2)
[[where t=ex]]