Find the Integral Using an Appropriate Trigonometric Substitution $\int \frac { x ^ { 3 } } { \sqrt { x ^ { 2 } + 36 } } d x$

Find the integral using an appropriate trigonometric substitution. $\int \frac { x ^ { 3 } } { \sqrt { x ^ { 2 } + 36 } } d x$

A) $\frac { 1 } { 3 } \left( x ^ { 2 } + 36 \right) ^ { 3 / 2 } \sqrt { x ^ { 2 } + 36 } + C$
B) $\frac { 1 } { 3 } \left( x ^ { 2 } + 72 \right) \sqrt { x ^ { 2 } + 36 } + C$
C) $\frac { 1 } { 3 } \left( x ^ { 2 } - 72 \right) \sqrt { x ^ { 2 } + 36 } + C$
D) $\frac { 1 } { 3 } \left( x ^ { 2 } - 36 \right) ^ { 3 / 2 } \sqrt { x ^ { 2 } + 36 } + C$

Correct Answer:

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Access For Free