What is the general method for integrals of type ∫ dx / (L√Q)?

For an integral of the form ∫ dx / (Linear√(Quadratic)), the standard substitution is to let the linear term L = 1/t. This transforms the radical into a form that can be integrated using standard formulas.

How do we simplify the quadratic expression 4x² + 8x + 3?

The quadratic can be written in terms of the linear factor (4x + 6) by using the substitution 4x + 6 = 1/t, or by completing the square as 4(x + 1)² - 1.

What is the derivative of the substitution 4x + 6 = 1/t?

Differentiating both sides with respect to x gives 4 dx = -1/t² dt, which implies dx = -dt / (4t²).

How is the constant of integration 'C' determined?

The value of C is found using the given initial condition I(0) = √3/4 + 20. After integrating, substitute x=0 and solve for C.

What does gcd(a, b) = 1 mean?

It means that the numbers a and b are coprime, implying the fraction a/b is in its simplest reduced form.

What is the integral of 1/√(a² - x²)?

The integral is sin⁻¹(x/a) + C. In this problem, a similar form arises after substitution.

Why is the substitution L = 1/t used specifically?

This substitution is effective because it reduces the degree of the terms under the square root, typically resulting in a pure quadratic in 't' that fits standard integration patterns.

Is the final answer an integer?

Yes, the question asks for a + b + c where these are natural numbers, and the sum results in 31.

What are common mistakes in this problem?

Common errors include incorrect differentiation during substitution, sign errors in the radical simplification, or failing to properly use the I(0) condition to find the correct value of c.

How much time should this problem take in JEE Main?

Given the substitution and the need to find the constant, this is a medium-difficulty problem that should take approximately 3-4 minutes.

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, c ∈ N, gcd(a, b) = 1, then a + b + c is equal to

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, c ∈ N, gcd(a, b) = 1, then a + b + c is equal to | JEE Main Mathematics

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Math Calculus

Q Numerical Integration

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

A) $30$ B) $29$ C) $28$ D) $31$

✅ Correct Answer

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Solution Steps

Standard Substitution for Linear-Quadratic Radical

For an integral of the form $\int \frac{dx}{L\sqrt{Q}}$, we let the linear part $L = \frac{1}{t}$.

Let $4x + 6 = \frac{1}{t} \implies 4dx = -\frac{1}{t^2}dt \implies dx = -\frac{dt}{4t^2}$.

Also, $4x = \frac{1}{t} – 6 \implies 2x = \frac{1}{2t} – 3$.

Transform the Quadratic expression

The quadratic is $4x^2 + 8x + 3 = (2x+2)^2 – 1$.

Substituting $2x = \frac{1}{2t} – 3$:

$((\frac{1}{2t} – 3) + 2)^2 – 1 = (\frac{1}{2t} – 1)^2 – 1 = \frac{1}{4t^2} – \frac{1}{t} + 1 – 1 = \frac{1 – 4t}{4t^2}$.

Thus, $\sqrt{4x^2+8x+3} = \sqrt{\frac{1-4t}{4t^2}} = \frac{\sqrt{1-4t}}{2t}$.

Substitute into the Integral

$I(x) = \int \frac{3 \cdot (-\frac{dt}{4t^2})}{(\frac{1}{t}) \cdot \frac{\sqrt{1-4t}}{2t}} = \int \frac{-\frac{3}{4t^2} dt}{\frac{\sqrt{1-4t}}{2t^2}} = -\frac{3}{2} \int \frac{dt}{\sqrt{1-4t}}$.

Integrating: $I(x) = -\frac{3}{2} \cdot \frac{\sqrt{1-4t}}{1/2 \cdot (-4)} + C = \frac{3}{4} \sqrt{1-4t} + C$.

Re-substitute for $x$

Since $t = \frac{1}{4x+6}$, we have $I(x) = \frac{3}{4} \sqrt{1 – \frac{4}{4x+6}} + C$.

$I(x) = \frac{3}{4} \sqrt{\frac{4x+6-4}{4x+6}} + C = \frac{3}{4} \sqrt{\frac{4x+2}{4x+6}} + C = \frac{3}{4} \sqrt{\frac{2x+1}{2x+3}} + C$.

Find Constant $C$ using $I(0)$

$I(0) = \frac{3}{4} \sqrt{\frac{0+1}{0+3}} + C = \frac{3}{4\sqrt{3}} + C = \frac{\sqrt{3}}{4} + C$.

Given $I(0) = \frac{\sqrt{3}}{4} + 20 \implies \frac{\sqrt{3}}{4} + C = \frac{\sqrt{3}}{4} + 20 \implies C = 20$.

Calculate $I(1/2)$ and find $a, b, c$

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

$I(1/2) = \frac{3\sqrt{2}}{8} + 20$.

Comparing with $\frac{a\sqrt{2}}{b} + c$: $a=3, b=8, c=20$.

$a+b+c = 3 + 8 + 20 = 31$

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Theory

1. Integration of Irrational Functions

The integration of functions containing square roots of algebraic expressions often requires specific substitutions to rationalize the integrand. When dealing with $\int \frac{dx}{(ax+b)\sqrt{px^2+qx+r}}$, the substitution $ax+b = 1/t$ is the most efficient method. This substitution simplifies the linear factor and converts the quadratic under the root into a new quadratic in $t$, which usually fits a standard integration form like $1/\sqrt{a^2-x^2}$ or $1/\sqrt{x^2 \pm a^2}$. Mastering these forms is crucial for JEE Main as they frequently appear in the calculus section.

2. Method of Substitution (u-substitution)

Integration by substitution is based on the chain rule for derivatives. If $g(x)$ is a differentiable function whose range is an interval $I$ and $f$ is continuous on $I$, then $\int f(g(x))g'(x) dx = \int f(u) du$. In this problem, choosing the correct $u$ (or $t$) involves identifying the part of the expression whose derivative is present (or can be easily manipulated into being present) in the numerator. The change of variables also requires updating the differentials correctly, as seen with $dx = -dt/4t^2$.

3. Determining Constants of Integration

An indefinite integral represents a family of curves, each differing by a constant $C$. To find a specific member of this family (a specific function $I(x)$), a boundary condition or initial value must be provided. In this problem, $I(0)$ acts as that condition. By plugging in $x=0$ into the general solution and equating it to the given value, we isolate the value of $C$. This step is critical in numerical-type problems where the final answer depends on the exact value of the function at a different point.

4. Coprime Numbers and GCD

A fraction $a/b$ is in its simplest form when the greatest common divisor (gcd) of $a$ and $b$ is 1. This means there are no common factors other than 1 between the numerator and denominator. In JEE problems, this condition ensures that the student provides a unique, irreducible set of values for $a$ and $b$. For example, if we had obtained $6/16$, we would have to reduce it to $3/8$ to satisfy $\text{gcd}(a,b)=1$ before calculating $a+b+c$.

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FAQs

Why did we use $4x+6 = 1/t$ and not just $x+1 = 1/t$?

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While $x+1$ is a factor related to the vertex of the quadratic, the substitution $L = 1/t$ is designed to specifically eliminate the linear factor $L$ outside the square root, making the radical easy to integrate.

What happens if the quadratic inside is $x^2 + 1$?

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The same substitution $L = 1/t$ still works. It will transform the expression into a form like $\int dt/\sqrt{t^2 \pm k}$, which results in a logarithmic (ln) or inverse hyperbolic result.

Can this problem be solved using trigonometric substitution?

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Yes, after completing the square in the quadratic, you could use secant or tangent substitution, but it is generally much more tedious than the $1/t$ method.

Is $a, b, c$ always positive in such questions?

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The question specifies $a, b, c \in \mathbf{N}$ (Natural numbers), which are positive integers $\{1, 2, 3, \dots\}$.

How do I verify if my integral result is correct?

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Differentiate your result $I(x)$ using the chain rule. It should return the original integrand.

What is $\text{gcd}(3, 8)$?

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Since 3 is prime and not a factor of 8, their greatest common divisor is 1. This satisfies the problem constraint.

Does the ‘3’ in the numerator change the substitution?

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No, it’s just a constant multiplier. It can be taken outside the integral to simplify calculations.

What if $1-4t$ was negative?

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Then the integral would involve complex numbers or the domain of $x$ would be restricted such that $1-4t \ge 0$. In JEE, the domain is usually valid for the given range.

How do you simplify $\sqrt{2/4}$ to $1/\sqrt{2}$?

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$\sqrt{2/4} = \sqrt{1/2} = \sqrt{1}/\sqrt{2} = 1/\sqrt{2}$. Then multiplying by 3/4 gives $3/(4\sqrt{2}) = 3\sqrt{2}/8$.

Is this a common JEE Main pattern?

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Yes, integrals of the form $1/(L\sqrt{Q})$ are a staple in the JEE calculus syllabus.

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