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

question 125

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

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