Evaluate $\int x \sqrt { x ^ { 2 } + 9 } d x$ .
A) $\frac { \left( x ^ { 2 } + 9 \right) ^ { 3 / 2 } } { 3 } + C$
B) $\left( x ^ { 2 } + 9 \right) ^ { 3 / 2 } + C$
C) $\frac { x ^ { 3 } } { 3 } + 9 x + C$
D) $\frac { 3 \left( x ^ { 2 } + 9 \right) ^ { 3 / 2 } } { 4 } + C$

Question

Quizplus · Accepted Answer

To solve the integral $\int x \sqrt { x ^ { 2 } + 9 } d x$, we can use a substitution method. Let $u = x^2 + 9$, then $du = 2x dx$. The integral becomes $\frac{1}{2} \int \sqrt{u} du$, which simplifies to $\frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C$, or $\frac{1}{3} (x^2 + 9)^{3/2} + C$, matching choice A.

Evaluate ∫xx2+9dx\int x \sqrt { x ^ { 2 } + 9 } d x∫xx2+9​dx

Evaluate $\int x \sqrt { x ^ { 2 } + 9 } d x$