Exercise 7.4 - Question 7
Problem
$$\int \frac{x-1}{\sqrt{x^2 - 1}} \, dx$$
Step-by-Step Solution
Step 1: Split the Integral
$$\int \frac{x-1}{\sqrt{x^2 - 1}} \, dx = \int \frac{x}{\sqrt{x^2 - 1}} \,
dx - \int \frac{1}{\sqrt{x^2 - 1}} \, dx$$
Step 2: Solve First Integral
For ∫x/√(x²-1) dx, let u = x² - 1, then du = 2x dx:
$$\int \frac{x}{\sqrt{x^2 - 1}} \, dx = \frac{1}{2}\int \frac{du}{\sqrt{u}}
= \frac{1}{2} \cdot 2\sqrt{u} = \sqrt{x^2 - 1}$$
Step 3: Solve Second Integral
The second is a standard form:
$$\int \frac{1}{\sqrt{x^2 - 1}} \, dx = \ln\left|x + \sqrt{x^2 - 1}\right| +
C$$
Step 4: Combine Results
$$= \sqrt{x^2 - 1} - \ln\left|x + \sqrt{x^2 - 1}\right| + C$$
✅ Final Answer
$$\int \frac{x-1}{\sqrt{x^2 - 1}} \, dx = \sqrt{x^2 - 1} - \ln\left|x +
\sqrt{x^2 - 1}\right| + C$$