<s>[INST] How to evaluate the integral $\int\frac{\sqrt{x^2-9}}{x^3}\;\mathrm d x$? $$\int\frac{\sqrt{x^2-9}}{x^3}\;\mathrm d x$$
The question is ask me to evaluate the integral but I have no idea how to start?
If there are any formulas required for this question, can you please list them ?
Thank you for any help!
 [/INST] Try this
$$ t^2 = x^2 - 9, \quad t\,dt = x\,dx $$
so
$$ \int \frac{\sqrt{x^2-9}}{x^3}\,dx = \int \frac{\sqrt{x^2-9}}{x^4} x\,dx \\
= \int \frac{t}{(t^2+9)^2} t\,dt \\= \int t \frac{t}{(t^2+9)^2} dt $$
Now do integration by parts
$$ u = t, \quad dv = \frac{t}{(t^2+9)^2}\,dt $$
$$ du = dt, \quad v = -\frac{1}{2(t^2+9)}$$
$$ \int \frac{t^2}{(t^2+9)^2} dt = -\frac{t}{2(t^2+9)} + \int \frac{1}{2(t^2+9)}\,dt \\
= -\frac{t}{2(t^2+9)} + \frac{1}{2\sqrt{3}} \arctan{\frac{t}{\sqrt{3}}} + C \\
= -\frac{\sqrt{x^2-9}}{2x^2} + \frac{1}{2\sqrt{3}} \arctan{\frac{\sqrt{x^2-9}}{\sqrt{3}}} + C $$
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