Let I(x)= 3dx(4x+6)(√4x^2+8x+3) and I(0)=√34+20.

Question

Let $I(x)=\int \frac{3dx}{(4x+6)(\sqrt{4{x}^{2}+8x+3})}$ and $I(0)=\frac{\sqrt{3}}{4}+20$. If

$I(\frac{1}{2})=\frac{a\sqrt{2}}{b}+c$, where $a,b,c\in N,gcd(a,b)=1$, then $a+b+c$ is equal to :

Accepted Answer

$I(x)=\int \frac{3dx}{(4x+6)\sqrt{4{x}^{2}+8x+3}}=\int \frac{3dx}{2(2x+3)\sqrt{(2x+2{)}^{2}-1}}$

let $2x+3=\frac{1}{t}\implies 2dx=-\frac{1}{{t}^{2}}dt\implies dx=-\frac{1}{2{t}^{2}}dt$

$\begin{gathered} 2x+2=\frac{1}{t}-1=\frac{1-t}{t} \ I(x)=\int \frac{3(-\frac{1}{2{t}^{2}}dt)}{2(\frac{1}{t})\sqrt{{(\frac{1-t}{t})}^{2}-1}}=\int \frac{-3dt}{4t\sqrt{\frac{1+{t}^{2}-2t-{t}^{2}}{{t}^{2}}}}=\int \frac{-3dt}{4\sqrt{1-2t}} \end{gathered}$

$\begin{gathered} I(x)=-\frac{3}{4}\int (1-2t{)}^{-1/2}dt \ I(x)=-\frac{3}{4}[\frac{(1-2t{)}^{1/2}}{(1/2)(-2)}]+C=\frac{3}{4}\sqrt{1-2t}+C \end{gathered}$

Substitute $t=\frac{1}{2x+3}$

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

$I(0)=\frac{\sqrt{3}}{4}+20\,$ (Given), substitute $x=0$ to find value of C.

$\frac{3}{4}\sqrt{\frac{1}{3}}+C=\frac{\sqrt{3}}{4}+20\implies \frac{\sqrt{3}}{4}+C=\frac{\sqrt{3}}{4}+20\implies C=20$

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

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

Comparing with $\frac{a\sqrt{2}}{b}+c$ :

$\begin{gathered} a=3,b=8,c=20 \ a+b+c=3+8+20=31 \end{gathered}$

Solutions · 1