Exercise 7.4 - Question 24 (MCQ)
Problem
$$\int \frac{dx}{x^2 + 2x + 2} \text{ equals}$$
(A) x tan⁻¹(x + 1) + C
(B) tan⁻¹(x + 1) + C
(C) (x + 1) tan⁻¹x + C
(D) tan⁻¹x + C
Step-by-Step Solution
Step 1: Complete the Square
$$x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x + 1)^2 + 1^2$$
Step 2: Rewrite the Integral
$$\int \frac{1}{(x+1)^2 + 1} \, dx$$
Step 3: Substitute
Let u = x + 1, then du = dx:
$$\int \frac{1}{u^2 + 1} \, du$$
Step 4: Apply Standard Formula
$$\int \frac{1}{u^2 + 1} \, du = \tan^{-1}(u) + C = \tan^{-1}(x + 1) + C$$
✅ Answer: (B) tan⁻¹(x + 1) + C
$$\int \frac{dx}{x^2 + 2x + 2} = \tan^{-1}(x + 1) + C$$