Exercise 7.4 - Question 10
Problem
$$\int \frac{1}{\sqrt{x^2 + 2x + 2}} \, dx$$
Step-by-Step Solution
Step 1: Complete the Square
$$x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x + 1)^2 + 1$$
Step 2: Rewrite the Integral
$$\int \frac{1}{\sqrt{(x+1)^2 + 1}} \, dx$$
Step 3: Substitute
Let u = x + 1, then du = dx:
$$\int \frac{1}{\sqrt{u^2 + 1}} \, du$$
Step 4: Apply Standard Formula
$$\int \frac{1}{\sqrt{u^2 + 1}} \, du = \ln\left|u + \sqrt{u^2 + 1}\right| +
C$$
Step 5: Substitute Back
$$= \ln\left|(x+1) + \sqrt{(x+1)^2 + 1}\right| + C$$
Simplify:
$$= \ln\left|(x+1) + \sqrt{x^2 + 2x + 2}\right| + C$$
✅ Final Answer
$$\int \frac{1}{\sqrt{x^2 + 2x + 2}} \, dx = \ln\left|(x+1) + \sqrt{x^2 + 2x
+ 2}\right| + C$$
Key Technique: Completing the Square
For any quadratic ax² + bx + c, complete the square to convert to (x + p)² ± q² form.