Match List-I with List-II

Choose the correct answer from the options given below:

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

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

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

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

$=\ln 2$

$(A)\rightarrow(III)$

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

$\sin^3 x$ 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).