Match List - I with List - II.

Choose the correct answer from the options given below:

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

A. $I=\int \frac{1}{x+\sqrt{x}} d x=\int \frac{1}{\sqrt{x}(\sqrt{x}+1)} d x$

let $y=(\sqrt{x}+1)$  so  $d y=\frac{1}{2 \sqrt{x}} d x$

so $2 d y=\frac{d x}{\sqrt{x}}$

$I=\int \frac{2 d y}{y} = 2 \log y + C$

$=2 \log (\sqrt{x}+1) + C$         →        III

B. $I=\int \frac{e^{\log \sqrt{x}}}{x} d x$

$=\int \frac{\sqrt{x}}{x} d x$

$=\int x^{-1 / 2} d x$ 

$=2\sqrt{x} + C$         →        I

C. $\int \frac{dx}{4x^2-9} = \frac{1}{4} \int \frac{dx}{x^2-(\frac{3}{2})^2}$

$= \frac{1 × 2}{2 × 4 × 3} \log \left|\frac{x - \frac{3}{2}}{x + \frac{3}{2}} \right|+ C$

$= \frac{1}{12} \log \left|\frac{2x - 3}{2x + 3}\right| + C$         →        II

D. $\int e^{\sqrt{x}} d x$   

so  $y=\sqrt{x} \Rightarrow y^2=x$

so $d x=2 y d y$

$\Rightarrow 2 \int y e^y d y=2\left[y e^y-\int e^y d y\right]$

$=2 y e^y-2 e^y+C$

$=2 e^y(y-1)+C$

$=2 e^{\sqrt{x}}(\sqrt{x}-1)+C$          →        II