Evaluate $\int x \sqrt { x ^ { 2 } + 4 } d x$ .
A) $\frac { 1 } { 3 } \left( x ^ { 2 } + 4 \right) ^ { 3 / 2 } + C$
B) $\left( x ^ { 2 } + 4 \right) ^ { 3 / 2 } + C$
C) $\frac { x ^ { 3 } } { 3 } + 4 x + C$
D) $\left( \frac { x ^ { 3 } } { 3 } + 4 \right) ^ { 3 / 2 } + C$

Question

Quizplus · Accepted Answer

To solve the integral $\int x \sqrt { x ^ { 2 } + 4 } d x$, we can use a substitution method. Let $u = x^2 + 4$, then $du = 2x dx$. The integral becomes $\frac{1}{2} \int \sqrt{u} du$, which evaluates to $\frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{3} (x^2 + 4)^{3/2} + C$, matching option A.

Evaluate ∫xx2+4dx\int x \sqrt { x ^ { 2 } + 4 } d x∫xx2+4​dx

Evaluate $\int x \sqrt { x ^ { 2 } + 4 } d x$