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

We have 
$$\int \frac{1}{\sqrt{1+x^2}}\, dx.$$
We make the following substitution. Let
$$
\begin{align*}
x&=\tan \theta \\
dx &= \sec^2 \theta \, d \theta\\
1+x^2&=1+\tan^2 \theta \\
&=\sec^2 \theta.
\end{align*}
$$
Hence our integral becomes
$$
\begin{align*}
\int \frac{1}{\sqrt{1+x^2}}\, dx &= \int\frac{\sec^2 \theta \, d \theta}{\sqrt{\sec^2 \theta}}\\
&=\int \sec \theta \, d \theta \\
&=\ln |\sec \theta + \tan \theta|+c.
\end{align*}
$$
For the back substitution, since
$$\tan \theta = \frac{x}{1},$$
we can form a right triangle with side opposite $\theta$ equal to $x$, and side adjacent to $\theta$ equal to $1$. Hence the hypoteneuse will have length $\sqrt{1+x^2}$. We can now read straight from the right triangle and back substitute,
\begin{align*}
\ln |\sec \theta + \tan \theta|+c &= \ln \left | \frac{\sqrt{1+x^2}}{1}+\frac{x}{1} \right |+c\\
&=\ln \left | \sqrt{1+x^2}+x \right |+c.
\end{align*}
