Question

1. $\int 5 x \sqrt3{3 x - 1} d x$

Explanation:

Step1: Use substitution

Let \(u = 3x-1\), then \(x=\frac{u + 1}{3}\) and \(dx=\frac{1}{3}du\).
The integral \(\int5x\sqrt[3]{3x - 1}dx\) becomes \(\int5\times\frac{u + 1}{3}\times u^{\frac{1}{3}}\times\frac{1}{3}du=\frac{5}{9}\int(u + 1)u^{\frac{1}{3}}du=\frac{5}{9}\int(u^{\frac{4}{3}}+u^{\frac{1}{3}})du\).

Step2: Integrate term - by - term

Using the power rule \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we have:
\(\frac{5}{9}(\frac{u^{\frac{4}{3}+1}}{\frac{4}{3}+1}+\frac{u^{\frac{1}{3}+1}}{\frac{1}{3}+1})+C=\frac{5}{9}(\frac{u^{\frac{7}{3}}}{\frac{7}{3}}+\frac{u^{\frac{4}{3}}}{\frac{4}{3}})+C\).

Step3: Substitute back \(u = 3x-1\)

\(\frac{5}{9}(\frac{3}{7}(3x - 1)^{\frac{7}{3}}+\frac{3}{4}(3x - 1)^{\frac{4}{3}})+C=\frac{5}{21}(3x - 1)^{\frac{7}{3}}+\frac{5}{12}(3x - 1)^{\frac{4}{3}}+C\).

Answer:

\(\frac{5}{21}(3x - 1)^{\frac{7}{3}}+\frac{5}{12}(3x - 1)^{\frac{4}{3}}+C\)