Match List-I with List-II List-I List-II (A) ∫limits_0^1(x^2)/(1 + x^…

The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

List-I	List-II
(A) $\int\limits_0^1\frac{x^2}{1 + x^3} dx$	(IV) $\frac{1}{3}\log_e 2$
B) $\int\limits_0^{\pi}3\sin x\, dx$	(III) 6
(C) $\int\limits_{-1}^1\sin^5x \cos^6 x\, dx$	(I) 0
(D) $\int\limits_2^3\frac{4}{x^2-1}dx$	(II) $2\log_e(\frac{3}{2})$

Evaluate each integral:

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

Let $x^3 = t$ ⟹ $3x^2 dx = dt$ ⟹ $\frac{1}{3} \int \frac{dt}{1 + t} = \frac{1}{3} \ln(1 + x^3)$

So, value = $\frac{1}{3} \ln(2)$⟹ matches (IV)

(B) $\int_0^\pi 3 \sin x \, dx = 3[-\cos x]_0^\pi = 3[-(-1) - (-1)] = 6$ ⟹ matches (III)

(C) $\int_{-1}^1 \sin^5 x \cos^6 x \, dx$ is an odd function (since $\sin^5 x$ is odd and $\cos^6 x$ is even). Hence integral over symmetric interval is 0 ⟹ matches (I)

(D)

Now integrate: \[ \int_{2}^{3} \left( \frac{2}{x - 1} - \frac{2}{x + 1} \right) dx = 2 \int_{2}^{3} \left( \frac{1}{x - 1} - \frac{1}{x + 1} \right) dx \]

Now integrate: $2 \int_{2}^3 \left( \frac{1}{x - 1} - \frac{1}{x + 1} \right) dx = 2[\ln|x - 1| - \ln|x + 1|]_{2}^3$

$= 2 \left( \ln\left(\frac{1}{2}\right) - \ln\left(\frac{1}{3}\right) \right) = 2 \ln\left(\frac{3}{2}\right)$⟹ matches (II)