Problem Statement
Evaluate the indefinite integral:
$\int \frac{1}{\sqrt{7 - 6x - x^2}} \, dx$
Verified Solution & Marking Scheme
Complete the Square Inside Radical
$7 - 6x - x^2 = 7 - (x^2 + 6x) = 7 - (x^2 + 6x + 9 - 9) = 7 - (x + 3)^2 + 9 = 16 - (x + 3)^2 = 4^2 - (x + 3)^2$
Apply Standard Formula ∫ 1/√(a² - u²) du
Let $u = x + 3 \implies du = dx$:
$\int \frac{1}{\sqrt{4^2 - u^2}} \, du = \arcsin\left(\frac{u}{4}\right) + C$
Substitute back $u = x + 3$:
$= \arcsin\left(\frac{x + 3}{4}\right) + C$