Question

Let \[ I(x) = \int \frac{3\,dx}{(4x+6)\sqrt{4x^2 + 8x + 3}} \] and \[ I(0) = \frac{\sqrt{3}}{4} + 20. \] If \[ I\left(\frac{1}{2}\right) = \frac{a\sqrt{2}}{b} + c, \] where \(a, b, c \in \mathbb{N}\) and \(\gcd(a,b)=1\), then find the value of \[ a + b + c. \]

• A. 29
• B. 28
• C. 30
• D. 31

Hint:

$\int \frac{dx}{(ax+b)\sqrt{px^2+qx+r}}$: Substitute $ax+b = 1/t$ or use trig substitution like $2x+2 = \sec\theta$.

Answer 1

The Correct Option is D

Rewrite the term inside the square root: $4x^2+8x+3 = (2x+2)^2 - 1$.
Let $2x+2 = \sec\theta$. Differentiating, $2dx = \sec\theta\tan\theta d\theta \implies dx = \frac{1}{2}\sec\theta\tan\theta d\theta$.
The term $4x+6 = 2(2x+2)+2 = 2\sec\theta+2 = 2(\sec\theta+1)$.
Substitute into the integral:
$I = \int \frac{3(\frac{1}{2}\sec\theta\tan\theta d\theta)}{2(\sec\theta+1)\sqrt{\sec^2\theta-1}} = \frac{3}{4} \int \frac{\sec\theta\tan\theta}{(\sec\theta+1)\tan\theta} d\theta$.
$I = \frac{3}{4} \int \frac{\sec\theta}{\sec\theta+1} d\theta = \frac{3}{4} \int \frac{1}{1+\cos\theta} d\theta$.
Using half-angle identity $1+\cos\theta = 2\cos^2(\theta/2)$:
$I = \frac{3}{4} \int \frac{1}{2\cos^2(\theta/2)} d\theta = \frac{3}{8} \int \sec^2(\theta/2) d\theta$.
$I = \frac{3}{8} [2\tan(\theta/2)] + C = \frac{3}{4}\tan(\theta/2) + C$.
Convert back to $x$: $\tan(\theta/2) = \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \sqrt{\frac{\sec\theta-1}{\sec\theta+1}} = \sqrt{\frac{2x+1}{2x+3}}$.
Given $I(0) = \frac{\sqrt{3}}{4} + 20$. At $x=0$, term is $\sqrt{1/3}$.
$\frac{3}{4}\frac{1}{\sqrt{3}} + C = \frac{\sqrt{3}}{4} + 20 \implies \frac{\sqrt{3}}{4} + C = \frac{\sqrt{3}}{4} + 20 \implies C=20$.
Find $I(1/2)$: At $x=1/2$, term is $\sqrt{\frac{2(0.5)+1}{2(0.5)+3}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$.
$I(1/2) = \frac{3}{4}\frac{1}{\sqrt{2}} + 20 = \frac{3\sqrt{2}}{8} + 20$.
Here $a=3, b=8, c=20$. $\gcd(3,8)=1$.
$a+b+c = 3+8+20 = 31$.