Match List-I with List-II

List-I	List-II
(A) $\int_0^1 \frac{x^2}{1 + x^3} dx$	(I) 0
(B) $\int_0^\pi 3\sin x dx$	(II) $2\log_e\left(\frac{3}{2}\right)$
(C) $\int_{-1}^1 \sin^5 x \cos^6 x dx$	(III) 6
(D) $\int_2^3 \frac{4}{x^2 - 1} dx$	(IV) $\frac{1}{3}\log_e 2$

Choose the correct answer from the options given below:

(A) $\int_0^1 \frac{x^2}{1 + x^3} dx$

Let $u = 1 + x^3$

$du = 3x^2 dx$

$x^2 dx = \frac{1}{3}du$

When $x = 0$: $u = 1 + 0^3 = 1$

When $x = 1$: $u = 1 + 1^3 = 2$

$\int_1^2 \frac{1}{3u} du$

$= \frac{1}{3}[\ln u]_1^2$

$= \frac{1}{3}(\ln 2 - \ln 1)$

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

Therefore (A) → (IV)

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(B) $\int_0^\pi 3\sin x dx$

$\int_0^\pi 3\sin x dx = 3[-\cos x]_0^\pi$

$= 3[(-\cos \pi) - (-\cos 0)]$

$= 3[(-(-1)) - (-1)]$

$= 3[1 + 1]$

$= 6$

Therefore (B) → (III)

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(C) $\int_{-1}^1 \sin^5 x \cos^6 x dx$

Check if the function is odd by replacing $x$ with $-x$:

$f(-x) = \sin^5(-x) \cos^6(-x)$

$= (-\sin x)^5 (\cos x)^6$

$= -\sin^5 x \cos^6 x$

$= -f(x)$

Since the function is odd and the interval $[-1, 1]$ is symmetric, the integral equals $0$.

Therefore (C) → (I)

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(D) $\int_2^3 \frac{4}{x^2 - 1} dx$

$\frac{4}{x^2 - 1} = \frac{4}{(x-1)(x+1)}$

Using partial fractions:

$\frac{4}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$

$4 = A(x+1) + B(x-1)$

Put $x = 1$: $4 = 2A$ so $A = 2$

Put $x = -1$: $4 = -2B$ so $B = -2$

$\int_2^3 \left(\frac{2}{x-1} - \frac{2}{x+1}\right) dx$

$= 2[\ln|x-1| - \ln|x+1|]_2^3$

$= 2\left[\ln\left|\frac{x-1}{x+1}\right|\right]_2^3$

$= 2\left[\ln\frac{2}{4} - \ln\frac{1}{3}\right]$

$= 2\left[\ln\frac{1}{2} - \ln\frac{1}{3}\right]$

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

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

Therefore (D) → (II)

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The correct matching is:

(A) → (IV)

(B) → (III)

(C) → (I)

(D) → (II)