Question

evaluate the integral by making the following substitution:
$\int x^{2}\sqrt{x^{3}+6} d x ; u=x^{3}+6$
$\bigcirc-\frac{2}{9}\left(x^{3}+6\
ight)^{\frac{1}{2}}+c$
$\bigcirc \frac{2}{9}\left(x^{3}+6\
ight)^{2}+c$
$\bigcirc \frac{2}{9}\left(x^{3}+6\
ight)^{\frac{3}{2}}+c$
$\bigcirc \frac{1}{9}\left(x^{3}+6\
ight)^{\frac{1}{2}}+c$
$\bigcirc$ the correct answer is not listed.

Explanation:

Step1: Differentiate \(u\)

Given \(u = x^{3}+6\), then \(du=3x^{2}dx\), so \(x^{2}dx=\frac{1}{3}du\).

Step2: Substitute into the integral

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

Step3: Integrate \(u\)

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

Step4: Simplify and substitute back \(u\)

\(\frac{1}{3}\cdot\frac{2}{3}u^{\frac{3}{2}}+C=\frac{2}{9}(x^{3}+6)^{\frac{3}{2}}+C\).

Answer:

\(\frac{2}{9}(x^{3}+6)^{\frac{3}{2}}+C\) (corresponds to the third option: \(\frac{2}{9}(x^{3}+6)^{\frac{3}{2}}+C\))