Let I ( x ) = 3 ? d x ( 4 x + 6 ) ( 4 x 2 + 8 x + 3 ) and I ( 0 ) = 3 4 + 20 . If I ( 1 2 ) = a 2 b + c , where a , b , c , gcd ( a , b ) = 1 , then a+b+c is equal to. [2026]

Question

Let $I (x) = \int \frac{3 d x}{(4 x + 6) (\sqrt{4 x^{2} + 8 x + 3})}$ and $I (0) = \frac{\sqrt{3}}{4} + 20 .$ If $I (\frac{1}{2}) = \frac{a \sqrt{2}}{b} + c, where a, b, c \in ℕ, \gcd (a, b) = 1,$ then $a + b + c$ is equal to. [2026]

Smart Achievers · Accepted Answer

(1)

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

$4 d x = - \frac{d t}{t^{2}},$ ${\frac{x + 1 = \frac{1}{t} - 2}{4}$

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

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

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

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

$= - \frac{3}{2} \int \frac{d t}{\sqrt{1 - 4 t}}$ $= - \frac{3}{2} (\frac{\sqrt{1 - 4 t}}{\frac{1}{2} \times - 4}) + C$

$= \frac{3}{4} \sqrt{1 - 4 t} + C$ $∵$ $t = \frac{1}{4 x + 6}$

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

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

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

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

$\Rightarrow C = 20$

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

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

$= \frac{3}{4 \sqrt{2}} + 20 = \frac{3 \sqrt{2}}{8} + 20$

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

Solution

Q.
Let $I (x) = \int \frac{3 d x}{(4 x + 6) (\sqrt{4 x^{2} + 8 x + 3})}$ and $I (0) = \frac{\sqrt{3}}{4} + 20 .$ If $I (\frac{1}{2}) = \frac{a \sqrt{2}}{b} + c, where a, b, c \in ℕ, \gcd (a, b) = 1,$ then $a + b + c$ is equal to. [2026]

1 31

2 30

3 29

4 28

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