The value of ∫_{e²}^{e⁴} 1/x ( e^{( (log_e x)² + 1 )⁻¹} / ( e^{( (log_e x)² + 1 )⁻¹} + e^{( (6 - log

Q: Wait, why do we change the limits when we substitute $t = \ln x$?

That is a fantastic question! When we perform a substitution, we are essentially changing the 'language' of the integral. The original limits $e^2$ and $e^4$ are values for $x$. Since our new integral is in terms of $t$, we must find the corresponding values of $t$ for those $x$ values. If $x = e^2$, then $t = \ln(e^2) = 2$. If $x = e^4$, then $t = \ln(e^4) = 4$. If you don't change the limits, you are mixing two different coordinate systems, which will lead to an incorrect result.

Q: Why does the King's Property work here? It feels like magic.

It does feel like magic, but it is actually rooted in symmetry! The King's Property, $\int_a^b f(t) dt = \int_a^b f(a+b-t) dt$, is essentially a reflection. When you replace $t$ with $(a+b-t)$, you are reflecting the function across the midpoint of the interval $[a, b]$. In this problem, the function is symmetric in a way that makes the denominator invariant under this transformation. It is not magic; it is the beautiful geometry of the function!

Q: Why can't we just integrate the function directly? It looks like a standard form.

I understand the frustration! The function $e^{(t^2+1)^{-1}}$ is not a standard elementary function that has a simple antiderivative. If you try to integrate it directly, you will get stuck in a loop of impossible integrals. The JEE Advanced examiners design these problems specifically so that you *cannot* integrate them directly. You are meant to use the properties of definite integrals to bypass the need for finding the antiderivative. Always look for properties first!

Question

The value of $\int_{e^{2}}^{e^{4}}\frac{1}{x}(\frac{e^{((\log_{e}x)^{2}+1)^{-1}}}{e^{((\log_{e}x)^{2}+1)^{-1}}+e^{((6-\log_{e}x)^{2}+1)^{-1}}})dx$ is

Options:

(A) 2

(B) $\log_{e}2$

(C) 1

(D) $\frac{1}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a massive, complex-looking integral. It is the kind of problem that makes your heart race, the kind that appears on a JEE Advanced paper and whispers, 'You cannot solve me.'

But today, we are going to dismantle that fear. Let us look at our problem:

$$I = \int_{e^{2}}^{e^{4}} \frac{1}{x} \left( \frac{e^{((\ln x)^{2}+1)^{-1}}}{e^{((\ln x)^{2}+1)^{-1}} + e^{((6-\ln x)^{2}+1)^{-1}}} \right) dx$$

At first glance, it is a nightmare of exponents and logarithms. But take a deep breath; in the world of calculus, complexity is often just a mask for elegance.

### The Key: Substitution

The first thing you should notice is the $\frac{1}{x}$ sitting outside the fraction. This is not a coincidence; it is a beacon. We know that the derivative of $\ln x$ is $\frac{1}{x}$.

Let us perform the substitution $t = \ln x$. When we differentiate both sides, we get $dt = \frac{1}{x} dx$.

This substitution is like finding the master key to a locked door. It absorbs the $\frac{1}{x}$ and transforms our integral into something much more manageable.

We must change our limits:
When $x = e^2$, $t = \ln(e^2) = 2$. 
When $x = e^4$, $t = \ln(e^4) = 4$.

Our integral now becomes:

$$I = \int_2^4 \frac{e^{(t^2+1)^{-1}}}{e^{(t^2+1)^{-1}} + e^{((6-t)^2+1)^{-1}}} dt$$

This is our **Equation 1**.

### The Mirror: King's Property

Now, look at the structure of the integrand. We have a fraction where the denominator is a sum of two terms that look like reflections of each other.

This is the signature of the **King's Property**:

$$\int_a^b f(t) dt = \int_a^b f(a+b-t) dt$$

Here, our limits are $a=2$ and $b=4$, so $a+b=6$. Let us apply this property by replacing every $t$ with $(6-t)$.

The integral becomes:

$$I = \int_2^4 \frac{e^{((6-t)^2+1)^{-1}}}{e^{((6-t)^2+1)^{-1}} + e^{((6-(6-t))^2+1)^{-1}}} dt$$

Look closely at the second term in the denominator: $(6-(6-t))^2 = t^2$. The sixes cancel out, and we are left with $t^2$.

Our integral is now:

$$I = \int_2^4 \frac{e^{((6-t)^2+1)^{-1}}}{e^{((6-t)^2+1)^{-1}} + e^{(t^2+1)^{-1}}} dt$$

This is **Equation 2**.

### The Grand Finale: Cancellation

Now, the magic happens. We have two expressions for $I$ (Equation 1 and Equation 2). They have the exact same denominator!

Let us add them together:

$$2I = \int_2^4 \frac{e^{(t^2+1)^{-1}} + e^{((6-t)^2+1)^{-1}}}{e^{(t^2+1)^{-1}} + e^{((6-t)^2+1)^{-1}}} dt$$

The numerator and denominator are identical. They cancel out to leave us with:

$$2I = \int_2^4 1 dt$$

This is the beauty of the JEE Advanced approach. We have reduced a terrifying, non-integrable function into the integral of a constant!

The integral of $1$ from $2$ to $4$ is simply $4-2 = 2$. So, $2I = 2$, which means **$I = 1$**.

We have conquered the monster. Remember, in these problems, the goal is never to fight the function head-on, but to find the symmetry that makes the problem collapse into simplicity.

Animated Solution for Mathematics - Definite Integration: The value of $\int_{e^{2}}^{e^{4}} \frac{1}{x} (\frac{e ^{((l o g_{e} x)^{2} + 1)^{- 1}}}{e ^{((l o g_{e} x)^{2} + 1)^{- 1}} + e ^{((6 - l o g_{e} x)^{2} + 1)^{- 1}}}) d x$ is

Select Answer:

Visualized Solution

Analyze the Integral Structure

Apply Substitution $t = lo g_{e} x$

Change the Limits of Integration

Rewrite Integral in terms of $t$

Introduce King's Property

Apply King's Property

Simplify the Transformed Integral

Add the Two Integrals

Simplify the Sum

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Key

The Mirror

The Grand Finale

Similar Questions

The value of $\int_{- 1/ 2}^{1/ 2} ((\frac{x + 1}{x - 1})^{2} + (\frac{x - 1}{x + 1})^{2} - 2)^{1/2} d x$ is:

The value of $\int_{- π /2}^{π /2} \frac{x ^{2} c o s x}{1 + e ^{x}} d x$ is equal to

The value of the integral $\int_{e^{- 1}}^{e^{2}} ∣ \frac{l o g _{e} x}{x} ∣ d x$ is

The value of $\int_{0}^{π} \frac{e ^{c o s x} s i n x}{( 1 + c o s ^{2} x ) ( e ^{c o s x} + e ^{- c o s x} )} d x$ is equal to

The value of the integral $\int_{- 2}^{2} \frac{∣ x ^{3} + x ∣}{( e ^{x ∣ x ∣} + 1 )} d x$ is equal to

The value of the integral $\int_{- 1}^{1} lo g_{e} (1 - x + 1 + x) d x$ is equal to :

The integral $\int_{1}^{e} {(\frac{x}{e})^{2 x} - (\frac{e}{x})^{x}} lo g_{e} x d x$ is equal to :

The value of the integral $\int_{- π /2}^{π /2} \frac{d x}{( 1 + e ^{x} ) ( s i n ^{6} x + c o s ^{6} x )}$ is equal to

The value of $\int_{- 1}^{1} \frac{( 1 + ∣ x ∣ - x ) e ^{x} + ( ∣ x ∣ - x ) e ^{- x}}{e ^{x} + e ^{- x}} d x$ is equal to

The value of $\int_{- π /2}^{π /2} (\frac{1 + s i n ^{2} x}{1 + π ^{s i n x}}) d x$ is

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 value of ∫e2e4​x1​(e((loge​x)2+1)−1+e((6−loge​x)2+1)−1e((loge​x)2+1)−1​)dx is