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+, where a, b, ∈ N, gcd(a, b)=1, then a+b+c is equal to

$4x^2+8x+3 = (2x+1)(2x+3)$4x2+8x+3=(2x+1)(2x+3) and $4x+6 = 2(2x+3)$4x+6=2(2x+3).

$\displaystyle I(x) = \frac{3}{4}\int \frac{du}{u^{3/2}\sqrt{u-2}}$I(x)=43​∫u3/2u−2​du​ where $u = 2x+3$u=2x+3.

Substituting $u = 2\sec^2\phi$u=2sec2ϕ: $I(x) = \frac{3}{4}\sin\phi + C = \frac{3}{4}\sqrt{\frac{2x+1}{2x+3}} + C$I(x)=43​sinϕ+C=43​2x+32x+1​​+C.

$I(0) = \frac{3}{4}\sqrt{\frac{1}{3}} + C = \frac{\sqrt{3}}{4} + C = \frac{\sqrt{3}}{4} + 20 \Rightarrow C = 20$I(0)=43​31​​+C=43​​+C=43​​+20⇒C=20.

$I\left(\frac{1}{2}\right) = \frac{3}{4}\sqrt{\frac{2}{4}} + 20 = \frac{3}{4\sqrt{2}} + 20 = \frac{3\sqrt{2}}{8} + 20$I(21​)=43​42​​+20=42​3​+20=832​​+20.

$a = 3, b = 8, c = 20$a=3,b=8,c=20, $\gcd(3,8) = 1$gcd(3,8)=1. $a+b+c = 31$a+b+c=31.