Match List-I with List-II

List-I	List-II
Definite integral	Value
(A) $\int_0^1 \frac{2x}{1 + x^2} dx$	(I) 2
(B) $\int_{-1}^1 sin^3 x \cos^4 x dx$	(II) $log_e\left(\frac{3}{2}\right)$
(C) $\int_0^{\pi} \sin x dx$	(III) $log_e 2$
(D) $\int_2^3 \frac{2}{x^2 - 1} dx$	(IV) 0

Choose the correct answer from the options given below:

For (A) $\int_0^1 \frac{2x}{1 + x^2} dx$:

Let $u = 1 + x^2$, then $du = 2x dx$

When $x = 0$: $u = 1$

When $x = 1$: $u = 2$

$\int_1^2 \frac{1}{u} du = [\log_e u]_1^2$

$= \log_e 2 - \log_e 1$

$= \log_e 2$

Therefore, (A) matches with (III)

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

The function $f(x) = \sin^3 x \cos^4 x$ is odd since $\sin^3 x$ is odd and $\cos^4 x$ is even.

For odd functions over symmetric intervals: $\int_{-a}^a f(x) dx = 0$

Therefore, (B) matches with (IV)

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For (C) $\int_0^{\pi} \sin x dx$:

$\int_0^{\pi} \sin x dx = [-\cos x]_0^{\pi}$

$= -\cos(\pi) - (-\cos(0))$

$= -(-1) + 1$

$= 2$

Therefore, (C) matches with (I)

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For (D) $\int_2^3 \frac{2}{x^2 - 1} dx$:

Using partial fractions: $\frac{2}{x^2-1} = \frac{2}{(x-1)(x+1)} = \frac{1}{x-1} - \frac{1}{x+1}$

$\int_2^3 \left(\frac{1}{x-1} - \frac{1}{x+1}\right) dx = \left[\log_e|x-1| - \log_e|x+1|\right]_2^3$

$= \left[\log_e\left|\frac{x-1}{x+1}\right|\right]_2^3$

$= \log_e\left(\frac{2}{4}\right) - \log_e\left(\frac{1}{3}\right)$

$= \log_e\left(\frac{3}{2}\right)$

Therefore, (D) matches with (II)

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The correct matching is: (A)-(III), (B)-(IV), (C)-(I), (D)-(II)