Evaluate $\int x \sqrt { x ^ { 2 } + 3 } d x$
A) $\left( x ^ { 2 } + 3 \right) ^ { 3 / 2 } + C$
B) $\frac { x ^ { 3 } } { 3 } + 3 x + C$
C) $\frac { 3 \left( x ^ { 2 } + 3 \right) ^ { 3 / 2 } } { 4 } + C$
D) $\frac { \left( x ^ { 2 } + 3 \right) ^ { 3 / 2 } } { 3 } + C$

Question

Quizplus · Accepted Answer

To solve the integral $\int x \sqrt { x ^ { 2 } + 3 } d x$, we can use a substitution method. Let $u = x^2 + 3$, then $du = 2x dx$. The integral becomes $\frac{1}{2} \int \sqrt{u} du$, which evaluates to $\frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{3} (x^2 + 3)^{3/2} + C$, matching option D.

Evaluate ∫xx2+3dx\int x \sqrt { x ^ { 2 } + 3 } d x∫xx2+3​dx

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