JEE Main 2026 — 28 January, Evening Shift — Question 19

$f(x)=\int \frac{d x}{x^{2 / 3}+2 x^{1 / 2}}$ Put $x=t^{6} \Rightarrow d x=6 t^{5} d t$

$=\int \frac{6 t^{5} d t}{t^{4}+2 t^{3}}=6 \int \frac{\left(t^{2}-4\right)+4}{t+2} d t$

$=6\left[\int(\mathrm{t}-2) \mathrm{dt}+4 \int \frac{1}{\mathrm{t}+2} \mathrm{dt}\right]$

$=6\left[\frac{\mathrm{t}^{2}}{2}-2 \mathrm{t}+4 \ell \mathrm{n}(\mathrm{t}+2)\right]+\mathrm{C}$

$=3 \mathrm{x}^{1 / 3}-12 \mathrm{x}^{1 / 6}+24 \ell \mathrm{n}\left(\mathrm{x}^{1 / 6}+2\right)+\mathrm{C}$

$\mathrm{f}(0)=24 \ell \mathrm{n} 2+\mathrm{C}=-26+24 \ell \mathrm{n} 2$ (given)

$\Rightarrow \mathrm{C}=-26$

Now $\mathrm{f}(1)=-35+24 \ell \mathrm{n} 3=\mathrm{a}+\mathrm{b} \ell \mathrm{n} 3$ (as given in ques.)

$\Rightarrow \mathrm{a}=-35 \& \mathrm{~b}=24$

$\Rightarrow \mathrm{a}+\mathrm{b}=-11$