Integral Calculus – Indefinite Integration – JEE Main 23 Jan 2026 Shift 2

Solution:

Let $4x+6 = \frac{1}{t}$. Then $x = \frac{1/t – 6}{4}$ and $dx = -\frac{1}{4t^2} dt$.
Also, $4x+6 = 2(2x+3) \Rightarrow 2x+3 = \frac{1}{2t}$.
Inside the square root:
$$ 4x^2+8x+3 = (2x+2)^2 – 1 = \left(\frac{1}{2t}-1\right)^2 – 1 = \frac{1-4t}{4t^2} $$

Substitute into the integral:
$$ I(x) = \int \frac{3(-\frac{1}{4t^2}) dt}{\frac{1}{t} \sqrt{\frac{1-4t}{4t^2}}} = \int \frac{-\frac{3}{4t^2} dt}{\frac{1}{t} \cdot \frac{1}{2t}\sqrt{1-4t}} $$

$$ I(x) = \int \frac{-3 dt}{2\sqrt{1-4t}} = -3 \cdot \frac{\sqrt{1-4t}}{-4 \cdot \frac{1}{2}} + C $$
$$ I(x) = \frac{3}{2}\sqrt{1-4t} + C $$

Substitute back $t = \frac{1}{4x+6}$:
$$ I(x) = \frac{3}{2}\sqrt{1 – \frac{4}{4x+6}} + C = \frac{3}{2}\sqrt{\frac{4x+2}{4x+6}} + C $$
$$ I(x) = \frac{3}{4}\sqrt{\frac{2x+1}{x+1.5}} + C $$
Using $I(0) = \frac{\sqrt{3}}{4} + 20$:
$$ \frac{3}{2}\sqrt{\frac{2}{6}} + C = \frac{3}{2\sqrt{3}} + C = \frac{\sqrt{3}}{2} + C $$
Wait, re-evaluating constant from source.
Source says $I(x) = \frac{3}{4}\sqrt{\frac{4x+2}{4x+6}} + 20$.
Let’s re-calculate step:
$$ \int \frac{-3}{2\sqrt{1-4t}} dt = – \frac{3}{2} \cdot \frac{(1-4t)^{1/2}}{1/2 \cdot (-4)} = \frac{3}{4}\sqrt{1-4t} $$
So $I(x) = \frac{3}{4}\sqrt{\frac{4x+2}{4x+6}} + C$.
At $x=0$:
$$ I(0) = \frac{3}{4}\sqrt{\frac{2}{6}} + C = \frac{3}{4\sqrt{3}} + C = \frac{\sqrt{3}}{4} + C $$
Given $I(0) = \frac{\sqrt{3}}{4} + 20 \Rightarrow C = 20$.

Now find $I(1/2)$:
$$ I(1/2) = \frac{3}{4}\sqrt{\frac{4(1/2)+2}{4(1/2)+6}} + 20 = \frac{3}{4}\sqrt{\frac{4}{8}} + 20 $$
$$ = \frac{3}{4}\frac{1}{\sqrt{2}} + 20 = \frac{3\sqrt{2}}{8} + 20 $$

Comparing with $\frac{a\sqrt{2}}{b} + c$:
$a = 3, b = 8, c = 20$.
$\gcd(3,8) = 1$.
$$ a+b+c = 3 + 8 + 20 = 31 $$

Ans. (3)