(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\neq - 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)