Reeves

2021-10-15

Find integral $\int \frac{\sqrt{{x}^{2}-9}}{x}dx$

Margot Mill

Step 1
Consider the integrals.
$\int \frac{\sqrt{{x}^{2}-9}}{x}dx$
Let,
$x=3\mathrm{sec}\left(t\right)$
Differentiating with respect to "t"
$dx=3\mathrm{sec}\left(t\right)\cdot \mathrm{tan}\left(t\right)dt$
Step 2
Substituting all value in given integrals,
$\int \frac{\sqrt{{x}^{2}-9}}{x}dx=\int \frac{\sqrt{9{\mathrm{sec}}^{2}t-9}}{3\mathrm{sec}\left(t\right)}3\mathrm{sec}\left(t\right)\cdot \mathrm{tan}\left(t\right)dt$
$=\int \frac{3\sqrt{{\mathrm{sec}}^{2}t-1}}{3\mathrm{sec}\left(t\right)}3\mathrm{sec}\left(t\right)\cdot \mathrm{tan}\left(t\right)dt$
$=\int 3\cdot \mathrm{tan}\left(t\right)\cdot \mathrm{tan}\left(t\right)dt$
$=3\int {\mathrm{tan}}^{2}\left(t\right)dt$
Further solving it,

$=3\left(\mathrm{tan}t-t\right)+C$
$=3\left[\mathrm{tan}\left({\mathrm{sec}}^{-1}\left(\frac{x}{3}\right)\right)-{\mathrm{sec}}^{-1}\left(\frac{x}{3}\right)\right]+C$
$=-3{\mathrm{sec}}^{-1}\left(\frac{x}{3}\right)+\sqrt{{x}^{2}-9}+C$

Do you have a similar question?