Question

$\int\frac{4x^{2}}{sqrt{3 - 16x^{3}}}dx=\square$

Explanation:

Step1: Use substitution

Let $u = 3-16x^{3}$, then $du=-48x^{2}dx$, and $4x^{2}dx=-\frac{1}{12}du$.

Step2: Rewrite the integral

The integral $\int\frac{4x^{2}}{\sqrt{3 - 16x^{3}}}dx$ becomes $\int\frac{-\frac{1}{12}du}{\sqrt{u}}=-\frac{1}{12}\int u^{-\frac{1}{2}}du$.

Step3: Integrate $u^{-\frac{1}{2}}$

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

Step4: Substitute back $u$

$-\frac{1}{12}\times2u^{\frac{1}{2}}+C=-\frac{1}{6}\sqrt{3 - 16x^{3}}+C$.

Answer:

$-\frac{1}{6}\sqrt{3 - 16x^{3}}+C$