Let I(x) = 3,dx(4x+6)√4x^2+8x+3.

Question

Let $I(x) = \displaystyle\int\frac{3\,dx}{(4x+6)\sqrt{4x^2+8x+3}}$. If $I(0) = \dfrac{\sqrt3}{4}+20$ and $I\left(\dfrac12\right) = \dfrac{a\sqrt2}{b}+c$, where $a,b,c\in\mathbb{N}$, $\gcd(a,b)=1$, then $a+b+c$ is equal to:

Accepted Answer

Write $4x^2+8x+3 = (2x+2)^2-1$, so $I(x)=\displaystyle\int\frac{3\,dx}{2(2x+3)\sqrt{(2x+2)^2-1}}$. Substitute $2x+3=1/t$; after simplification,

$$I(x) = \frac34\sqrt{1-\frac{2}{2x+3}}+C = \frac34\sqrt{\frac{2x+1}{2x+3}}+C$$

From $I(0)=\dfrac34\sqrt{\tfrac13}+C = \dfrac{\sqrt3}4+20$, we get $C=20$.

$$I\left(\frac12\right) = \frac34\sqrt{\frac24}+20 = \frac34\cdot\frac{\sqrt2}{2}+20 = \frac{3\sqrt2}{8}+20$$

So $a=3,\,b=8,\,c=20$, giving $a+b+c=31$.

Solutions · 1