Question

find the indefinite integral.

$$ int \frac { x } { sqrt { 3 x ^ { 2 } + 7 } } d x = square $$

$$ int \frac { x } { sqrt { 3 x ^ { 2 } + 7 } } d x $$

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 \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), 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:

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