JEE Advanced 2017 · Paper 2, Mathematics question 3

Substitute. Let $u=\sqrt{9+\sqrt x}$, so $du=\dfrac{dx}{4\sqrt x\,\sqrt{9+\sqrt x}}$. The equation becomes
$$dy=\frac{du}{2\sqrt{4+u}}\implies y=\sqrt{4+u}+C.$$

Constant. At $x=0$, $u=3$, so $y(0)=\sqrt7+C=\sqrt7$, giving $C=0$.

Value. At $x=256$, $u=\sqrt{9+16}=5$, so $y=\sqrt9=3$.

(A)