Question

evaluate the indefinite integral:
int left( \frac { 3 } { sqrt 3 { x } } - 6 sqrt 3 { x ^ { 2 } }
ight) d x = square + c.

Explanation:

Step1: Rewrite the integrand

Rewrite \(\frac{3}{\sqrt[3]{x}}\) as \(3x^{-\frac{1}{3}}\) and \(6\sqrt[3]{x^{2}}\) as \(6x^{\frac{2}{3}}\). So the integral becomes \(\int(3x^{-\frac{1}{3}}-6x^{\frac{2}{3}})dx\).

Step2: Integrate term - by - term

Use the power rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)).
For \(\int3x^{-\frac{1}{3}}dx\), we have \(3\times\frac{x^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}=3\times\frac{x^{\frac{2}{3}}}{\frac{2}{3}}=\frac{9}{2}x^{\frac{2}{3}}\).
For \(\int-6x^{\frac{2}{3}}dx\), we have \(-6\times\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}=-6\times\frac{x^{\frac{5}{3}}}{\frac{5}{3}}=-\frac{18}{5}x^{\frac{5}{3}}\).

Answer:

\(\frac{9}{2}x^{\frac{2}{3}}-\frac{18}{5}x^{\frac{5}{3}}\)