15 multiple choice 1 point ∫8/(x² - 4) dx = (a) 4tan⁻¹(x/2)+c (b) 8ln|x² - 4|+c (c) 2ln|(x - 2)/(x + 2)|+c…

Answer:

C. $2\ln\left|\frac{x - 2}{x + 2}\right|+C$

Explanation:

Step1: Decompose the fraction

We use partial - fraction decomposition. Given $\frac{8}{x^{2}-4}=\frac{8}{(x - 2)(x + 2)}=\frac{A}{x - 2}+\frac{B}{x + 2}$. Then $8=A(x + 2)+B(x - 2)$. Let $x = 2$, we get $8=A(2 + 2)+B(0)$, so $A = 2$. Let $x=-2$, we get $8=A(0)+B(-2 - 2)$, so $B=-2$. So $\frac{8}{x^{2}-4}=\frac{2}{x - 2}-\frac{2}{x + 2}$.

Step2: Integrate term - by - term

$\int\frac{8}{x^{2}-4}dx=\int\left(\frac{2}{x - 2}-\frac{2}{x + 2}\right)dx$. Using the integral formula $\int\frac{1}{u}du=\ln|u|+C$, we have $\int\frac{2}{x - 2}dx-\int\frac{2}{x + 2}dx=2\ln|x - 2|-2\ln|x + 2|+C$.

Step3: Use logarithm property

By the property $\ln a-\ln b=\ln\frac{a}{b}$, $2\ln|x - 2|-2\ln|x + 2|+C = 2\ln\left|\frac{x - 2}{x + 2}\right|+C$.