Question:

First, compute $f(x) = \int_1^x \frac{3t}{1+t^2} dt = \frac{3}{2}\ln(1+t^2)\big|_1^x = \frac{3}{2}\ln\frac{1+x^2}{2}$. For $0 < \alpha < \beta < 1$: since $f'(x) = \frac{3x}{1+x^2} > 0$, $f$ is strictly increasing, so $f(\alpha) < f(\beta)$ (eliminates option 2, confirms option 1). For $x \geq 1$, we need to check whether $f(x) + \frac{\pi}{4}$ compares with $\tan^{-1}x$. Define $g(x) = f(x) + \frac{\pi}{4} - \tan^{-1}x$. Then $g'(x) = \frac{3x}{1+x^2} - \frac{1}{1+x^2} = \frac{2x-1}{1+x^2} > 0$ for $x \geq 1$. Since $g(1) = \frac{3}{2}\ln 2 + \frac{\pi}{4} - \frac{\pi}{4} = \frac{3}{2}\ln 2 > 0$ and $g$ is increasing, $g(x) > 0$ for all $x \geq 1$ (confirms option 4, eliminates option 3).