The integral ∫₁ᵉ { (x/e)²ˣ - (e/x)ˣ } logₑ x dx is equal to : (1) 1/2 - e⁻¹ (2) 3/2 - e - 1/(2e²) (3

Q: Why did we choose $u = (\frac{x}{e})^x$ instead of just $u = x$ or $u = \ln x$?

Great question! In JEE Advanced problems, we look for the 'kernel' of the function—the part that, when differentiated, produces the rest of the integrand. Here, the term $(\frac{x}{e})^x$ is the 'heart' of the problem. Differentiating it yields a term involving $\ln x$, which is exactly what we have sitting outside the brackets. It's not a random choice; it's a strategic one designed to collapse the complexity.

Q: I always forget to change the limits of integration. Is it really necessary?

It is absolutely mandatory! If you don't change the limits, you are essentially mixing two different worlds: the $x$-world and the $u$-world. If you prefer to keep the original limits, you must integrate back to $x$ at the very end, which is often much harder. Changing the limits is the 'JEE way'—it keeps your math clean, elegant, and error-free.

Q: How did the $\log_e x$ term just vanish during the substitution?

It didn't vanish! It was 'absorbed' into the differential $du$. When we differentiated $\ln u = x(\ln x - 1)$, we found that $\frac{1}{u} du = \ln x dx$. So, the $\log_e x$ and the $dx$ combined perfectly to become part of the $du$ term. It's like a puzzle piece fitting perfectly into a slot.

Question

The integral $\int_1^e \left\{ \left(\frac{x}{e}\right)^{2x} - \left(\frac{e}{x}\right)^x \right\} \log_e x dx$ is equal to :

Options:

(A) $\frac{1}{2} - e^{-1}$

(B) $\frac{3}{2} - e - \frac{1}{2e^2}$

(C) $\frac{1}{2} + \frac{1}{e} - \frac{1}{2e^2}$

(D) $\frac{3}{2} - e - \frac{1}{2e^2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, student. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of exponents and logarithms. You see an integral like

$$I = \int_1^e \left\{ \left(\frac{x}{e}\right)^{2x} - \left(\frac{e}{x}\right)^x \right\} \log_e x \, dx$$

Your instinct might be to panic, but I want you to take a deep breath. In the world of JEE Advanced, complexity is often just a mask for a hidden, elegant simplicity. Our job is to peel back that mask.

### The Detective Work

Look at the integrand. We have $\left(\frac{x}{e}\right)^{2x}$ and $\left(\frac{e}{x}\right)^x$. Do you see the symmetry? The bases are reciprocals.

If we define $u = \left(\frac{x}{e}\right)^x$, then the first term is simply $u^2$, and the second term is $1/u$. This is the 'Aha!' moment. We aren't looking at a random collection of terms; we are looking at a structure waiting to be simplified.

### The Logarithmic Key

Now, how do we handle the $dx$? We need to find $du$. Since our substitution is $u = \left(\frac{x}{e}\right)^x$, we must use the power of logarithms.

Taking the natural log of both sides, we get $\ln u = x \ln \left(\frac{x}{e}\right)$. Using the properties of logarithms, this becomes $\ln u = x(\ln x - 1)$.

Now, differentiation becomes a breeze. Applying the product rule on the right side, we get:

$$\frac{d}{dx}(\ln u) = 1 \cdot (\ln x - 1) + x \cdot \frac{1}{x}$$

The magic happens here: the $-1$ and the $+1$ cancel out, leaving us with $\frac{1}{u} \frac{du}{dx} = \ln x$. This means $\frac{1}{u} du = \ln x \, dx$. The $\log_e x$ term that seemed so out of place is actually the key to our differential.

### The Transformation

We must now change our limits. When $x=1$, $u = (1/e)^1 = 1/e$. When $x=e$, $u = (e/e)^e = 1^e = 1$.

Our integral, which once spanned from $1$ to $e$ in the $x$-domain, now spans from $1/e$ to $1$ in the $u$-domain. The integral becomes:

$$I = \int_{1/e}^1 \left(u^2 - \frac{1}{u}\right) \frac{du}{u}$$

Distributing the $1/u$, we get:

$$I = \int_{1/e}^1 \left(u - \frac{1}{u^2}\right) du$$

### The Victory

We have arrived at the final stretch. Integrating $u$ gives us $\frac{u^2}{2}$, and integrating $-u^{-2}$ gives us $\frac{1}{u}$.

Evaluating this from $1/e$ to $1$, we calculate:

$$\left[ \frac{u^2}{2} + \frac{1}{u} \right]_{1/e}^1$$

Plugging in the upper limit, we get $\frac{1}{2} + 1 = \frac{3}{2}$. Plugging in the lower limit, we get:

$$\frac{(1/e)^2}{2} + \frac{1}{1/e} = \frac{1}{2e^2} + e$$

Subtracting the two, we arrive at our final, beautiful result:

$$\mathbf{\frac{3}{2} - e - \frac{1}{2e^2}}$$

See? The monster wasn't a monster at all. It was just a puzzle waiting for the right perspective. Keep this confidence with you—every complex integral is just a series of small, logical steps.

Animated Solution for Mathematics - Definite Integration: The integral $\int_{1}^{e} {(\frac{x}{e})^{2 x} - (\frac{e}{x})^{x}} lo g_{e} x d x$ is equal to :

Select Answer:

Visualized Solution

Analyze the Integrand

Define the Substitution $u$

Logarithmic Transformation

Differentiating the Equation

Simplifying the Derivative

Transforming the Integrand

Changing the Limits

Setting up the New Integral

Performing the Integration

Evaluating the Limits

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Detective Work

The Logarithmic Key

The Transformation

The Victory

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