The value of the integral ∫[-2 to 2] |x³ + x| / (e^(x|x|) + 1) dx is equal to (A) 5e², (B) 3e⁻², (C)

Q: Wait, why can't we just integrate this directly? It looks like a standard calculus problem.

I understand the urge to dive straight into integration, but look at the structure: $\frac{|x^3 + x|}{e^{x|x|} + 1}$. This is a non-elementary function. There is no standard antiderivative for this. In JEE Advanced, if an integral looks impossible to solve directly, it is almost always a signal to use a property—specifically, symmetry or the King's Property. We aren't meant to integrate it; we are meant to simplify it.

Q: How do I know when to use the $f(x) + f(-x)$ property?

Great question. The trigger is always the limits. If you see symmetric limits like $[-a, a]$ and the integrand is a complex fraction involving exponentials or trigonometric functions, this property is your best friend. It is designed to cancel out the 'ugly' parts of the function. If you see symmetric limits, always test $f(x) + f(-x)$ first.

Q: Why did the denominator just vanish? That feels like cheating.

It isn't cheating; it's the beauty of mathematical design! When you substitute $-x$ into the denominator $e^{x|x|} + 1$, you get $e^{-x|x|} + 1$. By rewriting $e^{-x|x|}$ as $\frac{1}{e^{x|x|}}$, you create a common denominator. When you add the two fractions, the numerator becomes $1 + e^{x|x|}$, which perfectly cancels the denominator. It is a classic JEE 'collapse' mechanism.

Question

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

Options:

(A) $5e^2$

(B) $3e^{-2}$

(C) $4$

(D) $6$

Sigma Percentile · Accepted Answer

### The Monster in the Integral

Imagine you are sitting in the exam hall. You turn the page, and there it is:

$$\int_{-2}^{2} \frac{|x^3 + x|}{(e^{x|x|} + 1)} dx$$

Your heart might skip a beat. It looks like a monster, featuring a modulus, a cubic polynomial, and an exponential function in the denominator.

If you try to integrate this directly, you will be lost in a labyrinth of cases and impossible substitutions. But here is the secret of the JEE Advanced: **The more intimidating the expression, the simpler the trick.**

### Phase 1: The Symmetric Insight

Look at the limits. They are symmetric: $[-2, 2]$. Whenever you see limits of the form $[-a, a]$, your brain should immediately pivot to the symmetry property of definite integrals.

We are not here to fight the function; we are here to transform it. We use the property:

$$\int_{-a}^{a} f(x) dx = \int_{0}^{a} [f(x) + f(-x)] dx$$

This property is like a folding mirror. It takes the area from the negative side and maps it onto the positive side, allowing us to combine the two into a single, manageable expression. Let us define our integrand as $f(x) = \frac{|x^3 + x|}{e^{x|x|} + 1}$.

### Phase 2: The Algebraic Dance

Now, we must find $f(-x)$. Let us look at the numerator first: $|(-x)^3 + (-x)|$.

Since the modulus absorbs the negative sign, this becomes $|-(x^3 + x)|$, which is simply $|x^3 + x|$. The numerator is an even function and stays exactly the same.

Now, consider the denominator: $e^{(-x)|-x|} + 1$. Since $|-x| = |x|$, this becomes $e^{-x|x|} + 1$. Now, let us combine $f(x)$ and $f(-x)$:

$$f(x) + f(-x) = |x^3 + x| \left[ \frac{1}{e^{x|x|} + 1} + \frac{1}{e^{-x|x|} + 1} \right]$$

Take a deep breath. Let $t = e^{x|x|}$. The bracketed expression becomes:

$$\frac{1}{t+1} + \frac{1}{t^{-1}+1} = \frac{1}{t+1} + \frac{t}{1+t} = \frac{1+t}{1+t} = 1$$

The entire exponential denominator has vanished!

### Phase 3: The Collapse

We are left with something beautiful. Our monstrous integral has collapsed into:

$$I = \int_{0}^{2} |x^3 + x| dx$$

Because our limits are now from $0$ to $2$, the value of $x$ is always positive. Therefore, $x^3 + x$ is always positive, and we can safely remove the modulus sign.

We are simply integrating a polynomial:

$$I = \int_{0}^{2} (x^3 + x) dx$$

### Phase 4: The Final Stretch

This is the part where you celebrate. We have reduced a terrifying exponential-modulus integral into a basic polynomial integration.

The antiderivative of $x^3$ is $\frac{x^4}{4}$, and the antiderivative of $x$ is $\frac{x^2}{2}$.

$$I = \left[ \frac{x^4}{4} + \frac{x^2}{2} \right]_{0}^{2}$$

Plugging in the upper limit $x=2$:

$$\frac{2^4}{4} + \frac{2^2}{2} = \frac{16}{4} + \frac{4}{2} = 4 + 2 = 6$$

The lower limit $x=0$ gives us $0$. The final answer is **6**.

### Conclusion

Do you see what happened? We didn't solve the problem by brute force; we solved it by understanding the structure.

The JEE Advanced is not a test of how much you can calculate, but of how well you can see the underlying patterns. Whenever you face a problem that looks impossible, stop. Look for the symmetry. Look for the cancellation. The answer is often hidden in plain sight, waiting for you to apply the right property.

Animated Solution for Mathematics - Definite Integration: The value of the integral $\int_{- 2}^{2} \frac{∣ x ^{3} + x ∣}{( e ^{x ∣ x ∣} + 1 )} d x$ is equal to

Select Answer:

Visualized Solution

Identifying Symmetric Limits

The Area to be Calculated

The Symmetric Integral Property

Setting up $f (- x)$

Evaluating the Numerator of $f (- x)$

Evaluating the Denominator of $f (- x)$

Summing $f (x)$ and $f (- x)$

Simplifying the Exponential Terms

The Magic of the Bracket

The Simplified Integral

Visualizing the New Area

Integrating the Polynomial

Substituting the Limits

Final Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Monster in the Integral

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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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 value of the integral ∫−22​(ex∣x∣+1)∣x3+x∣​dx is equal to