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
 A: 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.
A: 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*}
A: A possibility is to use the following properties of the inverse hyperbolic function $\operatorname{arcsinh}x$:


*

*$(\operatorname{arcsinh}x)' =\dfrac{1}{\sqrt{1+x^{2}}}\qquad$ (Wikipedia entry) 

*$\operatorname{arcsinh}x =\ln \left( x+\sqrt{x^{2}+1}\right)$ (Wikipedia entry)

A: Let $x=iy$, where $i=\sqrt{-1}$ . Then the integral becomes $\displaystyle\int\frac{i\cdot dy}{\sqrt{1-y^2}}=i\arcsin y=i\arcsin\frac xi$ $=i\cdot\arcsin(-ix)=i\cdot(-i)\cdot\text{arcsinh }x=\text{arcsinh }x=\ln(x+\sqrt{1+x^2})$ . See here and here for more details.
A: Let $\displaystyle 1+x^2=y^2\Rightarrow \sqrt{1+x^2} = y,$ and $2xdx = 2ydy$
$$\displaystyle xdx = ydy\Rightarrow \frac{dx}{y} = \frac{dy}{x} = \frac{d(x+y)}{(x+y)}$$ (Using Ratio and Proportion).
Now $$\displaystyle \int \frac{1}{\sqrt{1+x^2}}dx = \int \frac{dx}{y} = \int \frac{d(x+y)}{(x+y)} = \ln \left|x+y\right|+C$$
So $$\displaystyle \int \frac{1}{\sqrt{1+x^2}}dx = \ln \left|x+\sqrt{x^2+1}\right|+\mathbb{C}$$
