The integral ∫₂⁴ log x² / (log x² + log(36-12x+x²)) dx is equal to:
(a) 1
(b) 6
(c) 2
(d) 4

Q: Why did we rewrite $36-12x+x^2$ as $(6-x)^2$?

Great observation! In problems involving definite integrals with logarithmic terms, we are often looking for symmetry. By rewriting the quadratic as $(6-x)^2$, we reveal a beautiful relationship between the numerator and the denominator that becomes apparent only after applying the King's Property. It transforms a seemingly impossible integral into a simple one.

Q: Can we use the substitution $u = 6-x$ instead of the King's Property?

Absolutely! The King's Property is essentially a shortcut for that exact substitution. If you substitute $u = 6-x$, the limits change from $[2, 4]$ to $[4, 2]$, and the differential $dx$ becomes $-du$. When you flip the limits back to $[2, 4]$, you get the same result. The King's Property just saves you the extra steps of formal substitution.

Q: Why does adding the two integrals work so perfectly?

It works because the denominator is invariant under the transformation $x \to a+b-x$. When you replace $x$ with $6-x$, the term $\log x^2$ becomes $\log(6-x)^2$ and vice versa. Since addition is commutative, the denominator $\log x^2 + \log(6-x)^2$ remains unchanged. This allows us to add the numerators directly, which then perfectly match the denominator, leading to an integrand of $1$.

Question

The integral $\int_{2}^{4} \frac{\log x^2}{\log x^2 + \log(36-12x+x^2)} dx$ is equal to:

Options:

(A) 1

(B) 6

(C) 2

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a daunting mathematical structure: the integral

$$I = \int_{2}^{4} \frac{\log x^2}{\log x^2 + \log(36-12x+x^2)} dx$$

At first glance, it looks like a nightmare of logarithms and quadratic expressions. But in the world of JEE Advanced, intimidation is often just a mask for elegance.

Our goal is to find the area under this curve from $x=2$ to $x=4$. The first step in any such journey is to simplify the landscape.

Look closely at the denominator, specifically the term $36 - 12x + x^2$. If you rearrange the terms, you get $x^2 - 12x + 36$, which is the expansion of $(x-6)^2$, or equivalently $(6-x)^2$.

By making this simple substitution, our integral transforms into:

$$I = \int_{2}^{4} \frac{\log x^2}{\log x^2 + \log(6-x)^2} dx$$

Suddenly, the problem feels much more symmetric.

### The King's Property: Your Secret Weapon

Now that we have identified the symmetry, we need a tool to exploit it. Enter the **King's Property** of definite integrals:

$$\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$$

This property is a lifesaver in competitive exams. Here, our limits are $a=2$ and $b=4$, so $a+b = 6$.

The property tells us that if we replace every $x$ in our integrand with $(6-x)$, the value of the integral remains unchanged. Let's apply this transformation.

The numerator $\log x^2$ becomes $\log(6-x)^2$. The denominator, which was $\log x^2 + \log(6-x)^2$, transforms into $\log(6-x)^2 + \log(6-(6-x))^2$.

Simplifying the second term in the denominator, $6-(6-x)$ becomes $x$. Thus, our transformed integral is:

$$I = \int_{2}^{4} \frac{\log(6-x)^2}{\log(6-x)^2 + \log x^2} dx$$

### The Grand Cancellation: Where Magic Happens

We now have two expressions for the same integral $I$. Let's call the original one Equation (1) and the transformed one Equation (2).

If we add them together, we get:

$$2I = \int_{2}^{4} \frac{\log x^2}{\log x^2 + \log(6-x)^2} dx + \int_{2}^{4} \frac{\log(6-x)^2}{\log(6-x)^2 + \log x^2} dx$$

Because the denominators are identical, we can combine the numerators:

$$2I = \int_{2}^{4} \frac{\log x^2 + \log(6-x)^2}{\log x^2 + \log(6-x)^2} dx$$

Look at that! The numerator is exactly the same as the denominator. They cancel out perfectly to leave us with the integral of $1$ from $2$ to $4$:

$$2I = \int_{2}^{4} 1 dx$$

The integral of $1$ is simply $x$, and evaluating this from $2$ to $4$ gives us $4-2 = 2$. Therefore, $2I = 2$, which means **$I = 1$**.

We have conquered the beast! Remember, JEE problems are rarely about brute force; they are about spotting the hidden symmetry and choosing the right tool to reveal the beauty underneath.

Animated Solution for Mathematics - Definite Integration: The integral $\int_{2}^{4} \frac{l o g x ^{2}}{l o g x ^{2} + l o g ( 36 - 12 x + x ^{2} )} d x$ is equal to:

Select Answer:

Visualized Solution

Visualizing the Definite Integral

Simplifying the Denominator

Rewriting the Integral

The King's Property of Integration

Applying the Substitution

Simplifying the New Integral

Adding the Two Equations

Simplifying the Integrand

Evaluating the Integral

Finding the Final Value

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The King's Property

The Grand Cancellation

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral ∫24​logx2+log(36−12x+x2)logx2​dx is equal to: