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

Multiple Choice

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) $\frac{1}{3}\left(x^{2}+36\right) ^{3 / 2} \sqrt{x^{2}+36}+C$