Find $\int x ^ { 2 } \sqrt { x ^ { 3 } + 2 } d x$
A) $\frac { 2 } { 3 } \left( x ^ { 3 } + 2 \right) ^ { 1 / 2 } + C$
B) $\frac { 2 } { 9 } \left( x ^ { 3 } + 2 \right) ^ { 3 / 2 } + C$
C) $\frac { 2 } { 3 } \left( x ^ { 3 } + 2 \right) ^ { 3 / 2 } + C$
D) $\frac { 2 } { 9 } \left( x ^ { 3 } + 2 \right) ^ { 1 / 2 } + C$

Question

Quizplus · Accepted Answer

To solve the integral $\int x^2 \sqrt{x^3 + 2} dx$, let $u = x^3 + 2$, then $du = 3x^2 dx$. The integral becomes $\frac{1}{3} \int \sqrt{u} du = \frac{1}{3} \cdot \frac{2}{3} u^{3/2} + C = \frac{2}{9} (x^3 + 2)^{3/2} + C$.

Find ∫x2x3+2dx\int x ^ { 2 } \sqrt { x ^ { 3 } + 2 } d x∫x2x3+2​dx

Find $\int x ^ { 2 } \sqrt { x ^ { 3 } + 2 } d x$