# 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

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

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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.

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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*}

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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)
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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.

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

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