# How do you find the integral of (x)(secˉ¹(x)) dx ?

Apr 17, 2018

$\int x {\sec}^{-} 1 x \mathrm{dx} = \frac{x {\sec}^{-} 1 x}{2} - \frac{\sqrt{{x}^{2} - 1}}{2} + C$

#### Explanation:

Integrate by parts, making the following selections:

$u = {\sec}^{-} 1 \left(x\right)$

$\mathrm{du} = \frac{\mathrm{dx}}{x \sqrt{{x}^{2} - 1}}$

$\mathrm{dv} = x \mathrm{dx}$

$v = \frac{1}{2} {x}^{2}$

Then, apply the Integration by Parts formula:

$u v - \int v \mathrm{du} = \frac{{x}^{2} {\sec}^{-} 1 \left(x\right)}{2} - \frac{1}{2} \int \left({x}^{\left(\cancel{2}\right) 1} / \left(\left(\cancel{x}\right) \sqrt{{x}^{2} - 1}\right)\right) \mathrm{dx}$

$= \frac{x {\sec}^{-} 1 x}{2} - \frac{1}{2} \int \frac{x}{\sqrt{{x}^{2} - 1}} \mathrm{dx}$

$\int \frac{x}{\sqrt{{x}^{2} - 1}} \mathrm{dx}$ can be solved with a quick substitution:

$w = {x}^{2} - 1$

$\mathrm{dw} = 2 x \mathrm{dx}$

$x \mathrm{dx} = \frac{1}{2} \mathrm{dw}$

Thus, we have

$\frac{1}{2} \int \frac{1}{\sqrt{w}} \mathrm{dw} = \frac{1}{2} \int {w}^{- \frac{1}{2}} \mathrm{dw} = \cancel{\frac{1}{2}} \cancel{2} \sqrt{w} = \sqrt{{x}^{2} - 1}$

So, our integral is

$\int x {\sec}^{-} 1 x \mathrm{dx} = \frac{x {\sec}^{-} 1 x}{2} - \frac{\sqrt{{x}^{2} - 1}}{2} + C$