Question

(10) $\int\sqrt{3x + 5}dx=$

Explanation:

Step1: Use substitution

Let \(u = 3x+5\), then \(du=3dx\), and \(dx=\frac{1}{3}du\).
The integral becomes \(\int\sqrt{u}\cdot\frac{1}{3}du\).

Step2: Integrate \(\sqrt{u}\)

We know that \(\int u^{n}du=\frac{u^{n + 1}}{n+1}+C\) (\(n
eq - 1\)). For \(n=\frac{1}{2}\), \(\int\sqrt{u}du=\int u^{\frac{1}{2}}du=\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C=\frac{2}{3}u^{\frac{3}{2}}+C\).
So \(\frac{1}{3}\int\sqrt{u}du=\frac{1}{3}\times\frac{2}{3}u^{\frac{3}{2}}+C=\frac{2}{9}u^{\frac{3}{2}}+C\).

Step3: Substitute back \(u = 3x + 5\)

We get \(\frac{2}{9}(3x + 5)^{\frac{3}{2}}+C\).

Answer:

\(\frac{2}{9}(3x + 5)^{\frac{3}{2}}+C\)