∫8/√(12 - x² - 4x)dx = \n(a) 16√(12 - x² - 4x)+c \n(b) 2sin⁻¹((x + 2)/4)+c \n(c) 8sin⁻¹((x - 2)/4)+c \n(d)…

Explanation:

Step1: Complete the square for the denominator

First, rewrite the denominator $12 - x^{2}-4x$ as $12-(x^{2}+4x)$. Completing the square for $x^{2}+4x$ gives $(x + 2)^{2}-4$. So, $12-(x^{2}+4x)=12-((x + 2)^{2}-4)=16-(x + 2)^{2}$. Then the integral becomes $\int\frac{8}{\sqrt{16-(x + 2)^{2}}}dx$.

Step2: Use the integral formula for $\int\frac{1}{\sqrt{a^{2}-u^{2}}}du$

The integral formula is $\int\frac{1}{\sqrt{a^{2}-u^{2}}}du=\sin^{-1}(\frac{u}{a})+C$. In our case, $a = 4$ and $u=x + 2$, and we have a factor of 8 in the numerator. So, $\int\frac{8}{\sqrt{16-(x + 2)^{2}}}dx=8\int\frac{1}{\sqrt{4^{2}-(x + 2)^{2}}}dx$. Applying the formula, we get $8\sin^{-1}(\frac{x + 2}{4})+C$.

Answer:

D. $8\sin^{-1}(\frac{x + 2}{4})+C$