---

Solution

A. $\int \frac{1}{x+\sqrt{x}}dx$ Let $t=\sqrt{x}$ $x=t^2,\quad dx=2t\,dt$ Integral becomes $\int \frac{2t}{t^2+t}dt =2\int \frac{1}{t+1}dt =2\log(t+1)+C =2\log(\sqrt{x}+1)+C$ So A → III B. $\int \frac{e^{\log\sqrt{x}}}{x}dx$ Since $e^{\log\sqrt{x}}=\sqrt{x}$ $\int \frac{\sqrt{x}}{x}dx =\int x^{-1/2}dx =2\sqrt{x}+C$ So B → I C. $\int \frac{dx}{4x^2-9}$ Using standard formula $\int \frac{dx}{a^2x^2-b^2} =\frac{1}{2ab}\log\left|\frac{ax-b}{ax+b}\right|+C$ Here $a=2,\ b=3$ Result $\frac{1}{12}\log\left(\frac{2x-3}{2x+3}\right)+C$ So C → IV D. $\int e^{\sqrt{x}}dx$ Let $t=\sqrt{x}$ $x=t^2,\ dx=2t\,dt$ Integral $\int e^{\sqrt{x}}dx =2\int te^t dt$ Using integration by parts $2(t-1)e^t+C$ Substitute $t=\sqrt{x}$ $2(\sqrt{x}-1)e^{\sqrt{x}}+C$ So D → II