# How do you find the integral of sqrt(x^2-25)/x?

Mar 31, 2018

$\int \frac{\sqrt{{x}^{2} - 25}}{x} \mathrm{dx} = \sqrt{{x}^{2} - 25} - \arctan \left(\frac{\sqrt{{x}^{2} - 25}}{5}\right) + C$

#### Explanation:

The function is defined for $\left\mid x \right\mid \ge 5$. Restrict for the moment to $x > 5$ and substitute:

$x = 5 \sec t$

$\mathrm{dx} = 5 \sec t \tan t \mathrm{dt}$

with $t \in \left[0 , \frac{\pi}{2}\right)$.

Then:

$\int \frac{\sqrt{{x}^{2} - 25}}{x} \mathrm{dx} = 5 \int \frac{\sqrt{25 {\sec}^{2} t - 25} \sec t \tan t}{5 \sec t} \mathrm{dt}$

$\int \frac{\sqrt{{x}^{2} - 25}}{x} \mathrm{dx} = 5 \int \sqrt{{\sec}^{2} t - 1} \tan t \mathrm{dt}$

Use now the trigonometric identity:

${\sec}^{2} - 1 = {\tan}^{2} t$

and as for $\in \left[0 , \frac{\pi}{2}\right)$ the tangent is positive:

$\sqrt{{\sec}^{2} - 1} = \tan t$

so:

$\int \frac{\sqrt{{x}^{2} - 25}}{x} \mathrm{dx} = 5 \int {\tan}^{2} t \mathrm{dt}$

using the same identity again:

$\int \frac{\sqrt{{x}^{2} - 25}}{x} \mathrm{dx} = 5 \int \left({\sec}^{2} t - 1\right) \mathrm{dt}$

and for the linearity of the integral:

$\int \frac{\sqrt{{x}^{2} - 25}}{x} \mathrm{dx} = 5 \int {\sec}^{2} t \mathrm{dt} - 5 \int \mathrm{dt}$

$\int \frac{\sqrt{{x}^{2} - 25}}{x} \mathrm{dx} = 5 \tan t - 5 t + C$

To undo the substitution note that:

$\tan t = \sqrt{{\sec}^{2} t - 1} = \sqrt{{\left(\frac{x}{5}\right)}^{2} - 1} = \frac{1}{5} \sqrt{{x}^{2} - 25}$

and then:

$t = \arctan \left(\frac{\sqrt{{x}^{2} - 25}}{5}\right)$

So:

$\int \frac{\sqrt{{x}^{2} - 25}}{x} \mathrm{dx} = \sqrt{{x}^{2} - 25} - \arctan \left(\frac{\sqrt{{x}^{2} - 25}}{5}\right) + C$

By differentiating both sides we can verify that the solution extends also to $x < - 5$.