Integrate the function.
-$\int \frac { \sqrt { x ^ { 2 } - 9 } } { x } d x$
A) $\left[ \frac { \sqrt { x ^ { 2 } - 9 } } { 9 } - \sec ^ { - 1 } \left( \frac { x } { 3 } \right) \right] + C$
B) $3 \left[ \frac { \sqrt { x ^ { 2 } - 9 } } { 3 } - \sec ^ { - 1 } \left( \frac { x } { 3 } \right) \right] + C$
C) $3 \ln \left| \sqrt { x ^ { 2 } - 9 } - \left( \frac { x } { 3 } \right) \right| + C$
D) $3 \left[ \frac { \sqrt { x ^ { 2 } - 9 } } { 3 } - \sin ^ { - 1 } \left( \frac { x } { 3 } \right) \right] + C$

Question

Quizplus · Accepted Answer

To solve the integral, we can use trig substitution. Let $x=3\sec 	heta$, then $dx=3	an 	heta \sec 	heta d	heta$.
Substituting this in the integral and simplifying, we get:

\begin{align*}
\int \frac{\sqrt{x^2-9}}{x}dx &=\int \frac{\sqrt{(3\sec 	heta)^2-9}}{3\sec 	heta}3	an 	heta \sec 	heta d	heta \
&=\int \frac{\sqrt{9(\sec^2	heta-1)}}{3\sec 	heta}3	an 	heta \sec 	heta d	heta \
&=3\int 	an^2 	heta d	heta \
&=3\left[	an 	heta -	heta ight] +C \
&=3\left[\frac{\sqrt{x^2-9}}{3}-\sec^{-1}\left(\frac{x}{3}ight) ight]+C \
&=\sqrt{x^2-9}-3\sec^{-1}\left(\frac{x}{3}ight)+C
\end{align*}

Therefore, the best choice is $\boxed{	extbf{(B)}}$

Integrate the Function ∫x2−9xdx\int \frac { \sqrt { x ^ { 2 } - 9 } } { x } d x∫xx2−9​​dx

Integrate the Function $\int \frac { \sqrt { x ^ { 2 } - 9 } } { x } d x$