Exercise 7.4 - Question 12
Problem
$$\int \frac{1}{\sqrt{7 - 6x - x^2}} \, dx$$
Step-by-Step Solution
Step 1: Rearrange and Complete the Square
$$7 - 6x - x^2 = -(x^2 + 6x - 7) = -(x^2 + 6x + 9 - 9 - 7)$$
$$= -[(x + 3)^2 - 16] = 16 - (x + 3)^2$$
Step 2: Rewrite the Integral
$$\int \frac{1}{\sqrt{16 - (x+3)^2}} \, dx = \int \frac{1}{\sqrt{4^2 -
(x+3)^2}} \, dx$$
Step 3: Substitute u = x + 3
$$\int \frac{1}{\sqrt{16 - u^2}} \, du$$
Step 4: Apply Standard Formula
Using ∫1/√(a² - u²) du = sin⁻¹(u/a) with a = 4:
$$= \sin^{-1}\left(\frac{u}{4}\right) + C =
\sin^{-1}\left(\frac{x+3}{4}\right) + C$$
✅ Final Answer
$$\int \frac{1}{\sqrt{7 - 6x - x^2}} \, dx =
\sin^{-1}\left(\frac{x+3}{4}\right) + C$$