Question

evaluate the integral \\(\int x \sqrt3{36 + x^2} \\, dx\\).
\\(\int x \sqrt3{36 + x^2} \\, dx = \square\\)

Explanation:

Step1: Substitute \( u = 36 + x^{2} \)

Differentiate \( u \) with respect to \( x \): \( du=2x dx\), so \(x dx=\frac{1}{2}du\).
The integral \(\int x\sqrt[3]{36 + x^{2}}dx\) becomes \(\frac{1}{2}\int u^{\frac{1}{3}}du\).

Step2: Integrate \( u^{\frac{1}{3}} \)

Using the power - rule \(\int u^{n}du=\frac{u^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), for \(n=\frac{1}{3}\), we have \(\frac{1}{2}\times\frac{u^{\frac{1}{3}+1}}{\frac{1}{3}+1}+C=\frac{1}{2}\times\frac{u^{\frac{4}{3}}}{\frac{4}{3}}+C=\frac{3}{8}u^{\frac{4}{3}}+C\).

Step3: Substitute back \( u = 36 + x^{2} \)

We get \(\frac{3}{8}(36 + x^{2})^{\frac{4}{3}}+C\).

Answer:

\(\frac{3}{8}(36 + x^{2})^{\frac{4}{3}}+C\)