Let I(x) = ∫ 3dx / ((4x+6)(√(4x²+8x+3))) and I(0) = √3/4 + 20. If I(1/2) = (a√2)/b + c, where a, b,

Q: Why did we factor the quadratic inside the root? It looks like it could be completed as a square.

Completing the square is a valid strategy, but in this specific problem, factoring the quadratic $4x^2+8x+3$ into $(2x+1)(2x+3)$ is far more powerful. It reveals the common factor $(2x+3)$ that exists outside the root. This allows us to perform a clean substitution that simplifies the entire expression, rather than dealing with more complex inverse trigonometric forms.

Q: Why use $u = 2\sec^2\phi$ instead of the standard $u = 2\sec\phi$?

Great observation! If we used $u = 2\sec\phi$, the term inside the root would become $\sqrt{4\sec^2\phi - 4\sec\phi}$, which doesn't simplify nicely. By choosing $u = 2\sec^2\phi$, the term inside the root becomes $\sqrt{4\sec^4\phi - 4\sec^2\phi} = \sqrt{4\sec^2\phi(\sec^2\phi - 1)} = \sqrt{4\sec^2\phi \tan^2\phi} = 2\sec\phi \tan\phi$. This perfect square is exactly what we need to make the integral collapse beautifully.

Q: How do we handle the constant $C$ at the end?

The constant of integration $C$ is not just an arbitrary number; it is determined by the initial condition provided in the problem, $I(0) = \frac{\sqrt{3}}{4} + 20$. By substituting $x=0$ into our derived expression for $I(x)$ and equating it to the given value, the unknown constant $C$ is isolated and solved. Always remember to use the initial condition to find the specific solution for the given integral.

Question

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

Options:

(A) 30

(B) 31

(C) 29

(D) 28

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you first encounter an integral like $I(x) = \int \frac{3dx}{(4x+6)(\sqrt{4x^2+8x+3})}$, it is natural to feel a surge of intimidation. The combination of a linear term outside and a quadratic term inside a square root is a classic setup in JEE Advanced problems, designed to test your ability to see beyond the surface.

The secret here is not to rush into complex formulas, but to perform a bit of algebraic surgery first. Look at the quadratic term: $4x^2 + 8x + 3$. If we split the middle term, we get $4x^2 + 6x + 2x + 3$.

Factoring this, we find $2x(2x+3) + 1(2x+3)$, which gives us $(2x+1)(2x+3)$. Now, look at the linear term outside: $4x+6$. This is simply $2(2x+3)$. The presence of $(2x+3)$ in both places is not a coincidence; it is the path to the solution.

### The Substitution Strategy

With the common factor identified, we can rewrite our integral as $I(x) = \int \frac{3dx}{2(2x+3)\sqrt{(2x+1)(2x+3)}}$. This structure is practically begging for a substitution. Let $u = 2x+3$.

Then, $du = 2dx$, or $dx = \frac{du}{2}$. We also need to express the other factor, $2x+1$, in terms of $u$. Since $u = 2x+3$, it follows that $2x+1 = u-2$.

Substituting these into our integral, we get:

$$I(x) = \int \frac{3(\frac{du}{2})}{2u\sqrt{u(u-2)}} = \frac{3}{4} \int \frac{du}{u\sqrt{u^2-2u}}$$

This looks much cleaner, doesn't it? We have reduced a complex expression to a standard form that we can now tackle with a trigonometric pivot.

### The Trigonometric Pivot

Now we face $\int \frac{du}{u\sqrt{u^2-2u}}$. To resolve the square root, we need a substitution that turns $u^2-2u$ into a perfect square. Let $u = 2\sec^2\phi$.

Then $du = 4\sec^2\phi 	an\phi d\phi$. The term inside the root becomes $4\sec^4\phi - 4\sec^2\phi = 4\sec^2\phi(\sec^2\phi - 1) = 4\sec^2\phi 	an^2\phi$. The square root of this is $2\sec\phi 	an\phi$.

Substituting these back into the integral, we get:

$$\frac{3}{4} \int \frac{4\sec^2\phi 	an\phi d\phi}{(2\sec^2\phi)(2\sec\phi 	an\phi)} = \frac{3}{4} \int \frac{1}{\sec\phi} d\phi = \frac{3}{4} \int \cos\phi d\phi = \frac{3}{4} \sin\phi + C$$

Everything cancels out with such elegance! We are left with a simple integral of $\cos\phi$, which is $\sin\phi$.

### The Final Stretch

We must now return to our original variable $x$. Since $u = 2\sec^2\phi$, we have $\sec^2\phi = \frac{u}{2}$, which means $\cos^2\phi = \frac{2}{u}$.

Using the identity $\sin^2\phi = 1 - \cos^2\phi$, we find $\sin\phi = \sqrt{1 - \frac{2}{u}} = \sqrt{\frac{u-2}{u}}$. Substituting $u = 2x+3$ back in, we get $\sin\phi = \sqrt{\frac{2x+1}{2x+3}}$.

Thus, our integral is $I(x) = \frac{3}{4} \sqrt{\frac{2x+1}{2x+3}} + C$.

Using the initial condition $I(0) = \frac{\sqrt{3}}{4} + 20$, we find $C=20$. Finally, evaluating $I(\frac{1}{2})$ gives us $\frac{3\sqrt{2}}{8} + 20$.

Comparing this to $\frac{a\sqrt{2}}{b} + c$, we identify $a=3, b=8, c=20$. The sum **$a+b+c = 3+8+20 = 31$**. We have successfully navigated the complexity and arrived at the elegant solution.

Animated Solution for Mathematics - Indefinite Integration: Let $I (x) = \int \frac{3 d x}{( 4 x + 6 ) ( 4 x ^{2} + 8 x + 3 )}$ and $I (0) = \frac{3}{4} + 20$ . If $I (\frac{1}{2}) = \frac{a 2}{b} + c$ , where $a, b, c \in N, g cd (a, b) = 1$ , then $a + b + c$ is equal to

Select Answer:

Visualized Solution

Analyzing the Integrand

Factorizing the Quadratic Term

Rewriting the Linear Term

Reconstructing the Integral

First Substitution: $u = 2 x + 3$

Integral in Terms of $u$

Second Substitution: Trigonometric

Simplifying the Square Root

Integrating with Respect to $ϕ$

Converting Back to $x$

Finding the Constant $C$

Evaluating $I (\frac{1}{2})$

Final Comparison and Answer

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The Substitution Strategy

The Trigonometric Pivot

The Final Stretch

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Visualized Solution

Analyzing the Integrand

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Rewriting the Linear Term

Reconstructing the Integral

First Substitution: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u=2x+3

Integral in Terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u

Second Substitution: Trigonometric

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Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I(21​)