Exercise 7.4 - Question 22
Problem
$$\int \frac{x+3}{x^2 - 2x - 5} \, dx$$
Step-by-Step Solution
Step 1: Express Numerator
d/dx(x² - 2x - 5) = 2x - 2
Write x + 3 = A(2x - 2) + B
Solving: 2A = 1, so A = 1/2; and -2A + B = 3, so B = 4
Step 2: Split
$$= \frac{1}{2}\int \frac{2x-2}{x^2-2x-5} \, dx + 4\int \frac{1}{x^2-2x-5}
\, dx$$
Step 3: First Integral
$$I_1 = \frac{1}{2}\ln|x^2-2x-5|$$
Step 4: Second Integral
Complete the square: x² - 2x - 5 = (x-1)² - 6
$$I_2 = 4 \cdot
\frac{1}{2\sqrt{6}}\ln\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|$$
✅ Final Answer
$$= \frac{1}{2}\ln|x^2-2x-5| +
\frac{2}{\sqrt{6}}\ln\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right| + C$$