Substitute x = 3 sec u x=3\sec u x = 3 sec u 0 ≤ u < π 2 0\le u<\frac{\pi}{2} 0 ≤ u < 2 π d x = 3 sec u tan u d u \mathrm{d}x= 3\sec u\tan u\,\mathrm{d}u d x = 3 sec u tan u d u x 2 − 9 = 9 sec 2 u − 9 = 9 tan 2 u = 3 tan u \sqrt{x^2-9}=\sqrt{9\sec^2 u-9} =\sqrt{ 9\tan^2 u}=3\tan u x 2 − 9 = 9 sec 2 u − 9 = 9 tan 2 u = 3 tan u
∫ d x x 2 x 2 − 9 = ∫ 3 sec u tan u d u 9 sec 2 u 9 tan 2 u = 1 9 ∫ d u sec u = 1 9 ∫ cos u d u = 1 9 sin u + C . ! [ f i g u r e ] ( c l p / t i k z / c 7 a c b b d 75 e . p n g ) To evaluate sin u , we make a right triangle with angle u . Since sec u = x 3 = hyp adj , we label the hypotenuse x and the adjacent side 3 . Using the Pythagorean Theorem, the opposite side has length x 2 − 9 . So, sin u = x 2 − 9 x and ∫ d x x 2 x 2 − 9 = x 2 − 9 9 x + C . \begin{aligned}
\int \frac{\mathrm{d}x}{x^2\sqrt{x^2-9}}
&=\int \frac{3\sec u\tan u\ \mathrm{d}u}
{9\sec^2u\sqrt{9\tan^2u}} \\
&=\frac{1}{9}\int\frac{\mathrm{d}u}{\sec u} \\
&=\frac{1}{9}\int \cos u\ \mathrm{d}u
=\frac{1}{9}\sin u +C.\qquad \smash{
}
\\ \text{
To evaluate $\sin u$, we make a right triangle with angle $u$. Since $\sec u = \dfrac{x}{3} = \dfrac{\text{hyp}}{\text{adj}}$, we label the hypotenuse $x$ and the adjacent side $3$. Using the Pythagorean Theorem, the opposite side has length $\sqrt{x^2-9}$.
So, $\sin u = \dfrac{\sqrt{x^2-9}}x$ and} \\
\int \frac{\mathrm{d}x}{x^2\sqrt{x^2-9}} &= \frac{\sqrt{x^2-9}}{9x} +C.
\end{aligned} ∫ x 2 x 2 − 9 d x To evaluate sin u , we make a right triangle with angle u . Since sec u = 3 x = adj hyp , we label the hypotenuse x and the adjacent side 3. Using the Pythagorean Theorem, the opposite side has length x 2 − 9 . So, sin u = x x 2 − 9 and ∫ x 2 x 2 − 9 d x = ∫ 9 sec 2 u 9 tan 2 u 3 sec u tan u d u = 9 1 ∫ sec u d u = 9 1 ∫ cos u d u = 9 1 sin u + C . ! [ f i gu re ] ( c lp / t ik z / c 7 a c bb d 75 e . p n g ) = 9 x x 2 − 9 + C .
As a check, we observe that the derivative of the answer
d d x ( x 2 − 9 9 x + C ) = − x 2 − 9 9 x 2 + x 9 x x 2 − 9 = 1 9 − ( x 2 − 9 ) + x 2 x 2 x 2 − 9 = 1 x 2 x 2 − 9 \begin{aligned}
\frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\sqrt{x^2-9}}{9x}+C\right)
&=-\frac{\sqrt{x^2-9}}{9x^2} + \frac{x}{9x\sqrt{x^2-9}}
=\frac{1}{9}\ \frac{-(x^2-9)+x^2}{x^2\sqrt{x^2-9}} \\
&=\frac{1}{x^2\sqrt{x^2-9}}
\end{aligned} d x d ( 9 x x 2 − 9 + C ) = − 9 x 2 x 2 − 9 + 9 x x 2 − 9 x = 9 1 x 2 x 2 − 9 − ( x 2 − 9 ) + x 2 = x 2 x 2 − 9 1
is exactly the integrand. (We remark that this is the case even for x ≤ − 3 x\le -3 x ≤ − 3