Question

1. $\int 2x\sqrt{3x - 2}dx$

Explanation:

Step1: Substitution

Let \( u = 3x - 2 \), then \( x=\frac{u + 2}{3}\) and \( dx=\frac{1}{3}du\).
The integral becomes \(\int2\times\frac{u + 2}{3}\sqrt{u}\times\frac{1}{3}du=\frac{2}{9}\int(u + 2)u^{\frac{1}{2}}du=\frac{2}{9}\int(u^{\frac{3}{2}}+2u^{\frac{1}{2}})du\).

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:
\(\frac{2}{9}(\frac{u^{\frac{3}{2}+1}}{\frac{3}{2}+1}+2\times\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1})+C=\frac{2}{9}(\frac{2}{5}u^{\frac{5}{2}}+\frac{4}{3}u^{\frac{3}{2}})+C\).

Step3: Back - substitution

Substitute \( u = 3x - 2\) back:
\(\frac{4}{45}(3x - 2)^{\frac{5}{2}}+\frac{8}{27}(3x - 2)^{\frac{3}{2}}+C\).

Answer:

\(\frac{4}{45}(3x - 2)^{\frac{5}{2}}+\frac{8}{27}(3x - 2)^{\frac{3}{2}}+C\)