Match List-I with List-II

List-I (Definite integral)	List-II (Value)
(A) ${\int }_0^1\frac{2}{x}\mathrm{ d}x$	(I) 2
$$\begin{array}{r}{(B)\; \int }_{-1}^1{\sin }^{3}x{\cos }^{4}xdx\end{array}$$	(II) $\log \mathrm{<sub>e</sub>}\left(\frac{3}{2}\right)$
(C) ${\int }_0^{\pi }\sin xdx$	(III) $\log \mathrm{<sub>e</sub>}2$
(D) ${\int }_2^3\frac{2}{x^2-1}dx$	(IV) 0

Choose the correct answer from the options given below:

The correct option is (A) - (III), (B) - (IV), (C) - (I), (D) - (II).

Let us evaluate each definite integral in List-I step-by-step to find its corresponding value in List-II.

1. Evaluation of Integral (A):
The integral is:
$$I={\int }_0^1\frac{2x}{1+x^2}dx$$
To solve this, we can use the method of substitution. Let:
$$u=1+x^2$$
Differentiating both sides with respect to $x$ gives:
$$du=2xdx$$
Next, we determine the new limits of integration:
When $x=0$, $u=1+0^2=1$.
When $x=1$, $u=1+1^2=2$.
Substituting these values into the integral, we get:
$$I={\int }_1^2\frac{1}{u}du$$
Integrating $\frac{1}{u}$ yields:
$$I={\left[\log e_{}u\right]}_1^2=\log e_{}2-\log e_{}1$$
Since $\log e_{}1=0$, we have:
$$I=\log e_{}2$$
Thus, (A) matches with (III).

2. Evaluation of Integral (B):
The integral is:
$$I={\int }_{-1}^1{\sin }^{3}x{\cos }^{4}xdx$$
Let us analyze the integrand function:
$$f\left(x\right)={\sin }^{3}x{\cos }^{4}x$$
We test whether $f\left(x\right)$ is an odd or even function by replacing $x$ with $-x$:
$$f(-x)={\left(\sin \right(-x\left)\right)}^3{\left(\cos \right(-x\left)\right)}^4$$
Since $\sin (-x)=-\sin x$ and $\cos (-x)=\cos x$, we obtain:
$$f(-x)={(-\sin x)}^3{\left(\cos x\right)}^4=-{\sin }^{3}x{\cos }^{4}x=-f\left(x\right)$$
Because $f(-x)=-f\left(x\right)$, the integrand is an odd function.
By the properties of definite integrals, any integral of an odd function over symmetric limits $[-a,a]$ is zero:
$${\int }_{-a}^{a}f\left(x\right)dx=0$$
Therefore:
$$I=0$$
Thus, (B) matches with (IV).

3. Evaluation of Integral (C):
The integral is:
$$I={\int }_0^{\pi }\sin xdx$$
Evaluating the antiderivative of $\sin x$ gives:
$$I={[-\cos x]}_0^{\pi }=-\cos \pi -(-\cos 0)$$
Knowing that $\cos \pi =-1$ and $\cos 0=1$, we compute:
$$I=-(-1)+1=1+1=2$$
Thus, (C) matches with (I).

4. Evaluation of Integral (D):
The integral is:
$$I={\int }_2^3\frac{2}{x^2-1}dx$$
We use the standard integration formula:
$$\int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\log e_{}\left|\frac{x-a}{x+a}\right|$$
Setting $a=1$, our integral becomes:
$$I=2{\left[\frac{1}{2}\log e_{}\left|\frac{x-1}{x+1}\right|\right]}_2^3={\left[\log e_{}\left|\frac{x-1}{x+1}\right|\right]}_2^3$$
Now we apply the limits of integration:
$$I=\log e_{}\left(\frac{3-1}{3+1}\right)-\log e_{}\left(\frac{2-1}{2+1}\right)$$
Simplifying the fractions:
$$I=\log e_{}\left(\frac{2}{4}\right)-\log e_{}\left(\frac{1}{3}\right)=\log e_{}\left(\frac{1}{2}\right)-\log e_{}\left(\frac{1}{3}\right)$$
Using the logarithmic property $\log A-\log B=\log \left(\frac{A}{B}\right)$:
$$I=\log e_{}\left(\frac{1/2}{1/3}\right)=\log e_{}\left(\frac{3}{2}\right)$$
Thus, (D) matches with (II).

Conclusion:
By matching all items, we obtain:
(A) - (III), (B) - (IV), (C) - (I), (D) - (II).