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

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

Find the Integral Using an Appropriate Trigonometric Substitution ∫x3x2+36dx\int \frac{x^{3}}{\sqrt{x^{2}+36}} d x∫x2+36​x3​dx

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