find the indefinite integral.\n\n$$ int \frac { x } { sqrt { 3 x ^ { 2 } + 7 } } d x = square $$\n\n$$ int…

Explanation:

Step1: Use substitution

Let (u = 3x^{2}+7), then (du=6xdx), and (xdx=\frac{1}{6}du).

Step2: Substitute into the integral

(\int\frac{x}{\sqrt{3x^{2}+7}}dx=\int\frac{1}{\sqrt{u}}\cdot\frac{1}{6}du=\frac{1}{6}\int u^{-\frac{1}{2}}du).

Step3: Integrate using power rule

According to the power - rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), for (n=-\frac{1}{2}), (\frac{1}{6}\int u^{-\frac{1}{2}}du=\frac{1}{6}\cdot\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C).

Step4: Simplify the expression

(\frac{1}{6}\cdot\frac{u^{\frac{1}{2}}}{\frac{1}{2}}+C=\frac{1}{3}\sqrt{u}+C).

Step5: Back - substitute (u = 3x^{2}+7)

(\frac{1}{3}\sqrt{3x^{2}+7}+C).

Answer:

(\frac{1}{3}\sqrt{3x^{2}+7}+C)