Indefinite Integration
Trigonometric Substitution — Rational Integral
nta_pyq_2026_jan
Grade None
Question:
Let $I(x)=\displaystyle\int\frac{3\,dx}{(4z+6)\left(\sqrt{4x^2+8x+3}\right)}$ and $I(0)=\dfrac{\sqrt{3}}{4}+20$. If $I\!\left(\dfrac{1}{2}\right)=\dfrac{a\sqrt{2}}{b}+c$, where $a,b,c\in\mathbb{N}$, $\gcd(a,b)=1$, then $a+b+c$ is equal to
Step-by-Step Solution
Key Concept: $4x^2+8x+3=(2x+1)(2x+3)$ and $4x+6=2(2x+3)$. Let $u=2x+3$: $I=\tfrac{3}{4}\int\tfrac{du}{u^{3/2}\sqrt{u-2}}$. Substitute $u=2\sec^2\phi$: $I=\tfrac{3}{4}\sin\phi+C=\tfrac{3}{4}\sqrt{\tfrac{2x+1}{2x+3}}+C$.
$I(x)=\tfrac{3}{4}\sqrt{\tfrac{2x+1}{2x+3}}+20$. $a+b+c=31$.
Correct Answer: 2