2. ∫(x² + 2x)/√(x³ + 3x² + 7)dx\n(a) 2/3√(x³ + 3x² + 7)+c (b) 2/3√(3x² + 6x)+c (c) 1/2(√(x³ + 3x² + 7))+c…

Explanation:

Step1: Use substitution

Let $u = x^{3}+3x^{2}+7$, then $du=(3x^{2}+6x)dx = 3(x^{2}+2x)dx$, so $(x^{2}+2x)dx=\frac{1}{3}du$.

Step2: Rewrite the integral

The original integral $\int\frac{x^{2}+2x}{\sqrt{x^{3}+3x^{2}+7}}dx$ becomes $\frac{1}{3}\int u^{-\frac{1}{2}}du$.

Step3: Integrate

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\frac{1}{3}\times\frac{u^{\frac{1}{2}}}{\frac{1}{2}}+C=\frac{2}{3}\sqrt{u}+C$.

Step4: Substitute back

Substitute $u = x^{3}+3x^{2}+7$ back, we get $\frac{2}{3}\sqrt{x^{3}+3x^{2}+7}+C$.

Answer:

A. $\frac{2}{3}\sqrt{x^{3}+3x^{2}+7}+C$