Question

Evaluate \[ \int_0^1 \frac{x^2 - 2}{x^2 + 1} \, dx \]

• A. \( 1 + \frac{3\pi}{4} \)
• B. \( 1 - \frac{3\pi}{4} \)
• C. \( 1 - \frac{\pi}{4} \)
• D. \( 1 + \frac{\pi}{4} \)

Hint:

To integrate expressions involving \( \frac{1}{x^2 + 1} \), use the standard result \( \int \frac{1}{x^2 + 1} \, dx = \tan^{-1} x \).

Answer 1

The Correct Option is B

Step 1: Simplify the integral.
We start by splitting the integrand: \[ \frac{x^2 - 2}{x^2 + 1} = \frac{x^2 + 1 - 3}{x^2 + 1} = 1 - \frac{3}{x^2 + 1}. \] Thus, the integral becomes: \[ \int_0^1 \left( 1 - \frac{3}{x^2 + 1} \right) \, dx. \]
Step 2: Perform the integration.
The first term integrates to \( x \), and the second term is a standard integral, yielding \( \frac{3\pi}{4} \). So the total integral is: \[ \left[ x \right]_0^1 - 3 \left[ \tan^{-1} x \right]_0^1 = 1 - 3 \cdot \frac{\pi}{4} = 1 - \frac{3\pi}{4}. \]
Step 3: Conclusion.
Thus, the correct answer is \( 1 - \frac{3\pi}{4} \), corresponding to option (B).