Match List-I with List-II List-l Definite integral List-II Value (A)…

The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

List-l Definite integral	List-II Value
(A) $\int\limits_0^1\frac{2x}{1+x^2}dx$	(III) $\log_e2$
(B) $\int\limits_{-1}^1\sin^3x\cos^4x dx$	(IV) 0
(C) $\int\limits_0^{\pi}\sin x\,dx$	(I) 2
(D) $\int\limits_2^3\frac{2}{x^2-1}dx$	(II) $\log_e(\frac{3}{2})$

$(A)\ \int_{0}^{1}\frac{2x}{1+x^2}\,dx$

$=\left[\ln(1+x^2)\right]_{0}^{1}$

$=\ln 2$

$(A)\rightarrow(III)$

$(B)\ \int_{-1}^{1}\sin^3 x\cos^4 x\,dx$

$\sin^3 x$ is odd and $\cos^4 x$ is even, so the integrand is odd

$\int_{-1}^{1}\text{(odd function)}\,dx=0$

$(B)\rightarrow(IV)$

$(C)\ \int_{0}^{\pi}\sin x\,dx$

$=\left[-\cos x\right]_{0}^{\pi}$

$=(-\cos\pi)-(-\cos0)$

$=1+1=2$

$(C)\rightarrow(I)$

$(D)\ \int_{2}^{3}\frac{2}{x^2-1}\,dx$

$=\int_{2}^{3}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)dx$

$=\left[\ln|x-1|-\ln|x+1|\right]_{2}^{3}$

$=\ln\frac{2}{4}-\ln\frac{1}{3}$

$=\ln\frac{3}{2}$

$(D)\rightarrow(II)$

Final Matching: (A)-(III), (B)-(IV), (C)-(I), (D)-(II).