Question:

Step 1: Note that $\sin^{-1}\left(\dfrac{x}{\sqrt{1+x^2}}\right) = \tan^{-1}(x)$ and $\cos^{-1}\left(\dfrac{x}{\sqrt{1+x^2}}\right) = \dfrac{\pi}{2} - \tan^{-1}(x)$. Step 2: Compute $I_1 + I_2$: $$I_1 + I_2 = \int_0^1 \left[\tan^{-1}(x) + \frac{\pi}{2} - \tan^{-1}(x)\right]dx = \int_0^1 \frac{\pi}{2}\,dx = \frac{\pi}{2}.$$ Step 3: Compute $I_1 = \displaystyle\int_0^1 \tan^{-1}(x)\,dx$. Using integration by parts: $I_1 = \left[x\tan^{-1}(x)\right]_0^1 - \int_0^1 \dfrac{x}{1+x^2}dx = \dfrac{\pi}{4} - \dfrac{1}{2}\ln 2$. Step 4: Compute $I_2 = \dfrac{\pi}{2} - I_1 = \dfrac{\pi}{2} - \dfrac{\pi}{4} + \dfrac{1}{2}\ln 2 = \dfrac{\pi}{4} + \dfrac{1}{2}\ln 2$. Step 5: Compute $I_3 = \lim_{n\to\infty}\sum_{r=1}^{n}\dfrac{1}{n+r} = \int_0^1 \dfrac{1}{1+x}dx = \ln 2$. Step 6: Check option (a): $I_1 + I_3 = \dfrac{\pi}{4} - \dfrac{1}{2}\ln 2 + \ln 2 = \dfrac{\pi}{4} + \dfrac{1}{2}\ln 2 = I_2$ ✓. Step 7: Check option (d): $I_1 + I_2 + I_3 = \dfrac{\pi}{2} + \ln 2$ ✓. Step 8: Check option (b): $I_1 + I_2 + I_3 = \dfrac{\pi}{2} + \ln 2 \neq \ln 2$ ✗. Check option (c): $I_2 + I_3 = \dfrac{\pi}{4} + \dfrac{1}{2}\ln 2 + \ln 2 = \dfrac{\pi}{4} + \dfrac{3}{2}\ln 2 \neq I_1$ ✗. Correct options are (a) and (d).