How to integrate $\int{ dx \over \sqrt{1 + x^2}}$ How to integrate  $dx \over \sqrt{1 + x^2}$?
Answer should be $\log ( x + \sqrt{1 + x^2})$
Please help as possible...
Thank you

Let $\displaystyle I = \int \dfrac{1}{\sqrt{1+x^2}}~\mathrm dx.~ $Let $\tan u = x.$ Then  $\sec^2 u ~\mathrm du = \mathrm dx,~u = \arctan x$  and $$ \begin{align*}I  & = \int \sec u~ \mathrm du = \log (\tan u + \sec u)+C\\
& = \log\left(x + \sqrt{1 + x^2}\right) + C~.
\end{align*}$$
Feel free to ask questions if anything is unclear.
