Objective: Evaluate four definite integrals and match them to their corresponding values from List-II. This requires applying integration techniques including substitution, properties of odd/even functions, and partial fractions.
Integral (A) Evaluation: $\int_{0}^{1} \frac{2x}{1+x^2} dx$.
Using substitution $u = 1 + x^2$, $du = 2x dx$. The limits change from $x=0 \implies u=1$ and $x=1 \implies u=2$.
The integral becomes $\int_{1}^{2} \frac{1}{u} du = [\ln|u|]_{1}^{2} = \ln(2) - \ln(1) = \ln(2)$. This matches value (III).
Integral (B) Evaluation: $\int_{-1}^{1} \sin^3x \cos^4x dx$.
Let $f(x) = \sin^3x \cos^4x$. We check for odd/even properties:
$f(-x) = \sin^3(-x) \cos^4(-x) = (-\sin x)^3 (\cos x)^4 = -\sin^3x \cos^4x = -f(x)$.
Since $f(-x) = -f(x)$, the function is odd. The definite integral of an odd function over a symmetric interval $[-a, a]$ is 0. This matches value (IV).
Integral (C) Evaluation: $\int_{0}^{\pi} \sin x dx$.
The antiderivative of $\sin x$ is $-\cos x$.
$\int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi} = (-\cos(\pi)) - (-\cos(0)) = (-(-1)) - (-1) = 1 + 1 = 2$. This matches value (I).
Integral (D) Evaluation: $\int_{2}^{3} \frac{2}{x^2 - 1} dx$.
Using partial fraction decomposition: $\frac{2}{x^2 - 1} = \frac{2}{(x-1)(x+1)} = \frac{1}{x-1} - \frac{1}{x+1}$.
The integral is $\int_{2}^{3} (\frac{1}{x-1} - \frac{1}{x+1}) dx = [\ln|x-1| - \ln|x+1|]_{2}^{3} = [\ln|\frac{x-1}{x+1}|]_{2}^{3}$.
$= \ln|\frac{3-1}{3+1}| - \ln|\frac{2-1}{2+1}| = \ln|\frac{2}{4}| - \ln|\frac{1}{3}| = \ln(\frac{1}{2}) - \ln(\frac{1}{3}) = \ln(\frac{1/2}{1/3}) = \ln(\frac{3}{2})$. This matches value (II).
Conclusion: The correct pairings are (A) - (III), (B) - (IV), (C) - (I), and (D) - (II). This corresponds to option (4).