JEE Main 2026 — 23 January, Evening Shift — Question 14

Let $4 x+6=\frac{1}{t} \Rightarrow x=\frac{\frac{1}{t}-6}{4}$

$4 \mathrm{dx}=-\frac{\mathrm{dt}}{\mathrm{t}^{2}} \quad\left\{\frac{\mathrm{x}+1=\frac{1}{\mathrm{t}}-2}{4}\right.$

$\int \frac{3 d x}{(4 x+6) \sqrt{4(x+1)^{2}-1}}$

$\int \frac{3(-\mathrm{dt})}{4 \mathrm{t}^{2} \times \frac{1}{\mathrm{t}} \sqrt{\left(\frac{\frac{1}{\mathrm{t}}-2}{4}\right)^{2}-1}}$

$-\frac{3}{4} \int \frac{d t}{t \sqrt{\frac{(1-2 t)^{2}}{4 t^{2}}-1}}$

$-\frac{3}{4} \int \frac{\mathrm{dt}(2 \mathrm{t})}{\mathrm{t} \sqrt{1-4 \mathrm{t}}}$

$-\frac{3}{2} \int \frac{\mathrm{dt}}{\sqrt{1-4 \mathrm{t}}}=-\frac{3}{2}\left(\frac{\sqrt{1-4 \mathrm{t}}}{\frac{1}{2} \times-4}\right)+\mathrm{c}$

$=\frac{3}{4} \sqrt{1-4 \mathrm{t}}+\mathrm{c} \quad$

$\because \mathrm{t}=\frac{1}{4 \mathrm{x}+6}$

$=\frac{3}{4} \sqrt{1-4\left(\frac{1}{4 x+6}\right)}+c$

$=\frac{3}{4} \sqrt{\frac{4 x+6-4}{4 x+6}}+c$

$\mathrm{I}(\mathrm{x})=\frac{3}{4} \sqrt{\frac{4 \mathrm{x}+2}{4 \mathrm{x}+6}}+\mathrm{c}$

$\mathrm{I}(0)=\frac{3}{4} \sqrt{\frac{2}{6}}+\mathrm{c}$

$\mathrm{I}(0)=\frac{\sqrt{3}}{4}+\mathrm{c} \Rightarrow \mathrm{c}=20$

Hence $I(x)=\frac{3}{4} \sqrt{\frac{4 x+2}{4 x+6}}+20$

$\mathrm{I}\left(\frac{1}{2}\right)=\frac{3}{4} \sqrt{\frac{4}{8}}+20$

$=\frac{3}{4 \sqrt{2}}+20$

$=\frac{3 \sqrt{2}}{8}+20$

$a+b+c=3+8+20=31$