Exercise 7.4 - Question 11
Problem
$$\int \frac{1}{9x^2 + 6x + 5} \, dx$$
Step-by-Step Solution
Step 1: Complete the Square
$$9x^2 + 6x + 5 = 9\left(x^2 + \frac{2x}{3}\right) + 5$$
$$= 9\left(x^2 + \frac{2x}{3} + \frac{1}{9}\right) + 5 - 1 = 9\left(x +
\frac{1}{3}\right)^2 + 4$$
Step 2: Rewrite
$$\int \frac{1}{9(x + \frac{1}{3})^2 + 4} \, dx = \frac{1}{9}\int
\frac{1}{(x + \frac{1}{3})^2 + \frac{4}{9}} \, dx$$
Step 3: Let u = x + 1/3
$$= \frac{1}{9}\int \frac{1}{u^2 + (2/3)^2} \, du$$
Step 4: Apply Formula
Using ∫1/(u² + a²) du = (1/a)tan⁻¹(u/a) with a = 2/3:
$$= \frac{1}{9} \cdot \frac{3}{2}\tan^{-1}\left(\frac{3u}{2}\right) + C =
\frac{1}{6}\tan^{-1}\left(\frac{3x + 1}{2}\right) + C$$
✅ Final Answer
$$\int \frac{1}{9x^2 + 6x + 5} \, dx =
\frac{1}{6}\tan^{-1}\left(\frac{3x+1}{2}\right) + C$$