Let I(x)=intfrac{3dx}{(4x+6)(sqrt{4x^{2}+8x+3 | vedclass

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(C) Given $I(x) = \int \frac{3dx}{(4x+6)\sqrt{4x^2+8x+3}}$.Rewrite the integral as $I(x) = \int \frac{3dx}{(4x+6)\sqrt{(2x+2)^2-1}}$.Let $2x+2 = \sec

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$31$

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(C) Given $I(x) = \int \frac{3dx}{(4x+6)\sqrt{4x^2+8x+3}}$.Rewrite the integral as $I(x) = \int \frac{3dx}{(4x+6)\sqrt{(2x+2)^2-1}}$.Let $2x+2 = \sec 	heta$,then $2dx = \sec 	heta 	an 	heta d	heta$.Also,$4x+6 = 2(2x+2)+2 = 2\sec 	heta + 2 = 2(\sec 	heta + 1)$.Substituting these into the integral:$I(x) = \int \frac{3 \cdot \frac{1}{2} \sec 	heta 	an 	heta d	heta}{2(\sec 	heta + 1)	an 	heta} = \frac{3}{4} \int \frac{\sec 	heta}{\sec 	heta + 1} d	heta = \frac{3}{4} \int \frac{1}{\cos 	heta + 1} d	heta = \frac{3}{4} \int \frac{1}{2 \cos^2(	heta/2)} d	heta = \frac{3}{8} \int \sec^2(	heta/2) d	heta$.$I(x) = \frac{3}{8} \cdot 2 	an(	heta/2) + C = \frac{3}{4} 	an(	heta/2) + C$.Since $\sec 	heta = 2x+2$,$	an^2(	heta/2) = \frac{1-\cos 	heta}{1+\cos 	heta} = \frac{1-1/(2x+2)}{1+1/(2x+2)} = \frac{2x+1}{2x+3}$.So,$I(x) = \frac{3}{4} \sqrt{\frac{2x+1}{2x+3}} + C$.Given $I(0) = \frac{3}{4} \sqrt{\frac{1}{3}} + C = \frac{\sqrt{3}}{4} + C = \frac{\sqrt{3}}{4} + 20$,so $C=20$.Thus,$I(x) = \frac{3}{4} \sqrt{\frac{2x+1}{2x+3}} + 20$.For $x = 1/2$,$I(1/2) = \frac{3}{4} \sqrt{\frac{1+1}{1+3}} + 20 = \frac{3}{4} \sqrt{\frac{2}{4}} + 20 = \frac{3}{4} \cdot \frac{\sqrt{2}}{2} + 20 = \frac{3\sqrt{2}}{8} + 20$.Here $a=3, b=8, c=20$. $gcd(3,8)=1$.Therefore,$a+b+c = 3+8+20 = 31$.

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

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