integrals...i m lacking in the basics

post the complete solution dude..!

\cot^{-1} e^x = \tan^{-1} e^{-x}

ok sir and wat about the first one?

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=e^x]]