$\text { Use integration tables to find } \int \frac { 1 } { x ^ { 2 } \sqrt{ 36 x ^ {2} - 25} } d x \text {. }$
A) $\pm \frac { \sqrt { 36 x ^ { 2 } \pm 25 } } { 25 x } + C$
B) $\frac { \sqrt { 36 x ^ { 2 } - 25 } } { 36 x } + C$
C) $\frac { \sqrt { 36 x ^ { 2 } + 25 } } { 25 x } + C$
D) $- \frac { \sqrt { 36 x ^ { 2 } - 25 } } { 25 x } + C$
E) $\frac { \sqrt { 36 x ^ { 2 } - 25 } } { 25 x } + C$

Question

$\text { Use integration tables to find } \int \frac { 1 } {  x ^ { 2 } \sqrt{  36 x ^ {2} - 25} } d x \text {. }$
A) $\pm \frac { \sqrt { 36 x ^ { 2 } \pm 25 } } { 25 x } + C$
B) $\frac { \sqrt { 36 x ^ { 2 } - 25 } } { 36 x } + C$
C) $\frac { \sqrt { 36 x ^ { 2 } + 25 } } { 25 x } + C$
D) $- \frac { \sqrt { 36 x ^ { 2 } - 25 } } { 25 x } + C$
E) $\frac { \sqrt { 36 x ^ { 2 } - 25 } } { 25 x } + C$

Quizplus · Accepted Answer

We can use the substitution $u=\sqrt{36x^2-25}$. Then $du=\frac{36x}{\sqrt{36x^2-25}}dx=\frac{36x}{u}dx$, which implies $dx=\frac{u}{36x}du$. Substituting these into the given integral gives:
\begin{align*}
\int \frac{1}{x^2\sqrt{36x^2-25}}dx &= \int\frac{1}{x^2u}\cdot\frac{u}{36x}du\
&= \frac{1}{36}\int\frac{1}{xu}du\
&= \frac{1}{36}\int\frac{1}{x\sqrt{36x^2-25}}(36x)du\
&= \frac{1}{36}\int\frac{1}{u}du\
&= \frac{1}{36}\ln|u|+C\
&= \frac{1}{36}\ln|36x^2-25|^{1/2}+C\
&= \frac{\sqrt{36x^2-25}}{36x}+C.
\end{align*}
Therefore, the answer is E.

Use integration tables to find ∫1x236x2−25dx. \text { Use integration tables to find } \int \frac { 1 } { x ^ { 2 } \sqrt{ 36 x ^ {2} - 25} } d x \text {. } Use integration tables to find ∫x236x2−25​1​dx.

$\text { Use integration tables to find } \int \frac { 1 } { x ^ { 2 } \sqrt{ 36 x ^ {2} - 25} } d x \text {. }$