Evaluate the integral in terms of natural logarithms.
-$$\int _ { 4 } ^ { 7 } \frac { d x } { \sqrt { x ^ { 2 } - 9 } }$$
A) $\ln \left( \frac { 7 + \sqrt { 40 } } { 23 } \right) - 1$
B) $\frac { 1 } { 2 } \ln \left( \frac { \sqrt { 40 } } { 4 + \sqrt { 7 } } \right)$
C) $\ln 4$
D) $\ln \left( \frac { 7 + \sqrt { 40 } } { 4 + \sqrt { 7 } } \right)$

Question

Quizplus · Accepted Answer

We can begin by letting $x=3\sec	heta$. Then we have $dx=3\sec	heta	an	heta d	heta$ and $\sqrt{x^2-9}=3	an	heta$. Substituting yields:
\begin{align*}
\int_4^7\dfrac{dx}{\sqrt{x^2-9}}&=\int_{\sec^{-1}(4/3)}^{\sec^{-1}(7/3)}\dfrac{3\sec	heta	an	heta d	heta}{3	an	heta} \
&=\int_{\sec^{-1}(4/3)}^{\sec^{-1}(7/3)}\sec	heta d	heta \
&=\ln\left|\sec	heta+	an	hetaight|_{\sec^{-1}(4/3)}^{\sec^{-1}(7/3)} \
&=\ln\left|\dfrac{7}{3}+\dfrac{4}{3}\sqrt{2}ight|-\ln\left|\dfrac{4}{3}+\sqrt{7}ight|
\end{align*}
This simplifies to $\ln\left(\dfrac{7+\sqrt{40}}{4+\sqrt{7}}ight)$. Thus, the answer is $\boxed{	extbf{(D)}}$.

Evaluate the Integral in Terms of Natural Logarithms ∫47dxx2−9\int _ { 4 } ^ { 7 } \frac { d x } { \sqrt { x ^ { 2 } - 9 } }∫47​x2−9​dx​

Evaluate the Integral in Terms of Natural Logarithms $\int _ { 4 } ^ { 7 } \frac { d x } { \sqrt { x ^ { 2 } - 9 } }$