Analyzing the Setup
Welcome, fellow traveler of the mathematical landscape. Today, we stand before an integral that, at first glance, might seem like a tangled mess of exponentials and polynomials:
It is easy to feel overwhelmed by the (1+xex)2 in the denominator. But remember, in JEE Advanced, complexity is often just a mask for elegance. Let's peel back that mask together.
First, we must train our eyes to see the hidden relationships. Look at the term xex. What happens when we differentiate it?
Using the product rule, we get ex+xex, which is ex(x+1). This is the 'spark' of our solution. We have (x+1) in the numerator, but we are missing the ex.
The Algebraic Bridge
Forcing the Pattern
We don't just stare at the problem; we manipulate it. By multiplying the numerator and denominator by ex, we create the perfect differential for our substitution.
Now, the integral becomes:
This is the turning point. By setting t=1+xex, we transform the entire expression into a rational function of t.
The numerator ex(x+1)dx becomes dt, and the denominator becomes (t−1)t2. We have successfully moved from the world of exponentials to the world of algebra.
The Art of Decomposition
Breaking Down Barriers
Now, we use partial fraction decomposition. We break the expression into simpler components:
This is the beauty of calculus—taking a complex, intimidating structure and decomposing it into simple, solvable parts. Integrating these terms is straightforward:
∫(t−11−t1−t21)dt=log∣t−1∣−log∣t∣+t1+C
Final Calculation
Finally, we substitute back t=1+xex to return to our original variable x.
The result is:
This is not just an answer; it is a testament to your ability to see through the noise. Keep practicing, keep questioning, and remember that every integral is a story waiting to be told.