# How to integrate $\int{ dx \over \sqrt{1 + x^2}}$ [closed]

How to integrate $dx \over \sqrt{1 + x^2}$? Answer should be $\log ( x + \sqrt{1 + x^2})$

Thank you

-

## closed as off-topic by Bruno Joyal, user1337, Old John, Thomas, Carl MummertDec 18 '13 at 22:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Bruno Joyal, user1337, Old John, Thomas, Carl Mummert
If this question can be reworded to fit the rules in the help center, please edit the question.

Make the substitution $u=\sqrt{1+x^2}$ –  dreamer Dec 17 '13 at 18:30
$x = \sinh t$ works well. –  Daniel Fischer Dec 17 '13 at 18:30
Please use parentheses for grouping sqrt 1 + x^2 could easily be $\sqrt 1 + x^2$ instead of $\sqrt {1+x^2}$. Better, use MathJax. A tutorial is here –  Ross Millikan Dec 17 '13 at 18:41
One can learn how to recognize when an integral begs for a trigonometric substitution. That would have gotten you started here. (That's not necessarily the only way to do it, though.) –  Michael Hardy Dec 17 '13 at 18:48
This is a very standard integral and you could have differentiated $\log(x+\sqrt{1+x^{2}})$ and got the result. –  Indrayudh Roy Dec 17 '13 at 19:35

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.

-
Thanks you its clear now.. –  user116434 Dec 17 '13 at 18:57

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

1. $(\operatorname{arcsinh}x)' =\dfrac{1}{\sqrt{1+x^{2}}}\qquad$ (Wikipedia entry)
2. $\operatorname{arcsinh}x =\ln \left( x+\sqrt{x^{2}+1}\right)$ (Wikipedia entry)
-

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.

-

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 $$\displaystyle \int \frac{1}{\sqrt{1+x^2}}dx = \ln \left|x+\sqrt{x^2+1}\right|+\mathbb{C}$$

-