To solve this matching problem, we need to evaluate each definite integral in Column I and find its corresponding value in Column II.

Integral (A)

$I_A = \int\limits_{ - 1}^1 {{{dx} \over {1 + {x^2}}}}$

1. Find the antiderivative: The standard integral of $\frac{1}{1+x^2}$ is $\arctan(x)$.
2. Apply the Fundamental Theorem of Calculus: $I_A = [\arctan(x)]_{ - 1}^1 = \arctan(1) - \arctan(-1)$
3. Evaluate: We know $\arctan(1) = \frac{\pi}{4}$ and $\arctan(-1) = -\frac{\pi}{4}$. $I_A = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{2\pi}{4} = \frac{\pi}{2}$
4. Match: The value $\frac{\pi}{2}$ corresponds to (S) in Column II. Thus, A → S.

Integral (C)

$I_C = \int\limits_2^3 {{{dx} \over {1 - {x^2}}}}$

1. Find the antiderivative: The standard integral of $\frac{1}{1-x^2}$ is $\frac{1}{2}\ln\left|\frac{1+x}{1-x}\right|$.
2. Apply the Fundamental Theorem of Calculus: $I_C = \left[\frac{1}{2}\ln\left|\frac{1+x}{1-x}\right|\right]_2^3 = \frac{1}{2}\left(\ln\left|\frac{1+3}{1-3}\right| - \ln\left|\frac{1+2}{1-2}\right|\right)$
3. Evaluate: $I_C = \frac{1}{2}\left(\ln\left|\frac{4}{-2}\right| - \ln\left|\frac{3}{-1}\right|\right) = \frac{1}{2}(\ln(2) - \ln(3)) = \frac{1}{2}\ln\left(\frac{2}{3}\right)$
4. Match: Assuming $\log$ denotes the natural logarithm, this value corresponds to (P) in Column II. Thus, C → P.

Integral (D)

$I_D = \int\limits_1^2 {{{dx} \over {x\sqrt {{x^2} - 1} }}}$

1. Find the antiderivative: The standard integral of $\frac{1}{x\sqrt{x^2-1}}$ is $\operatorname{arcsec}(x)$.
2. Apply the Fundamental Theorem of Calculus: $I_D = [\operatorname{arcsec}(x)]_1^2 = \operatorname{arcsec}(2) - \operatorname{arcsec}(1)$
3. Evaluate: We know $\operatorname{arcsec}(2) = \arccos(\frac{1}{2}) = \frac{\pi}{3}$ and $\operatorname{arcsec}(1) = \arccos(1) = 0$. $I_D = \frac{\pi}{3} - 0 = \frac{\pi}{3}$
4. Match: This value corresponds to (R) in Column II. Thus, D → R.

Summary of reliable matches:

• A → S
• C → P
• D → R

Looking at the options, both B and C fit these matches. They only differ in the matching for integral (B).

• Option B: A-s, B-q, C-p, D-r
• Option C: A-s, B-s, C-p, D-r

Integral (B)

$I_B = \int\limits_0^1 {{{dx} \over {\sqrt {1 + {x^2}} }}}$

1. Find the antiderivative: The standard integral of $\frac{1}{\sqrt{1+x^2}}$ is $\ln(x + \sqrt{1+x^2})$.
2. Apply the Fundamental Theorem of Calculus: $I_B = [\ln(x + \sqrt{1+x^2})]_0^1 = \ln(1 + \sqrt{1+1^2}) - \ln(0 + \sqrt{1+0^2})$
3. Evaluate: $I_B = \ln(1+\sqrt{2}) - \ln(1) = \ln(1+\sqrt{2})$
4. Match: The value $\ln(1+\sqrt{2})$ does not match either of the possible values for B from the options:
   • (Q) $2\log(\frac{2}{3}) = 2\ln(\frac{2}{3}) = \ln(\frac{4}{9})$, which is negative, while $\ln(1+\sqrt{2})$ is positive.
   • (S) $\frac{\pi}{2} \approx 1.57$, while $\ln(1+\sqrt{2}) \approx 0.88$.

This indicates a likely typo in the question for Integral (B). A common typo is a sign error. Let's consider if the integral was meant to be: $I_B' = \int\limits_0^1 {{{dx} \over {\sqrt {1 - {x^2}} }}}$

1. Find the antiderivative: The antiderivative is $\arcsin(x)$.
2. Evaluate: $I_B' = [\arcsin(x)]_0^1 = \arcsin(1) - \arcsin(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$ This value matches (S) in Column II. Assuming this correction, we have B → S.

Final Matching

With the correction for integral (B), the complete matching is:

• A → S
• B → S
• C → P
• D → R

This corresponds to option C.