Animated Solution for Mathematics - Indefinite Integration: Let $f(x)=\frac{x}{(1+x^n{)}^{1/n}}$ for $n\ge 2$ and $g(x)=(f\circ f\circ ...\circ f)(x)$ ($f$ occurs $n$ times). Then $\int x^{n-2}g(x)dx$ equals

Welcome, fellow traveler on the path of JEE mastery! Today, we are going to unravel a problem that looks like a tangled mess of functions but is actually a beautifully choreographed dance of algebra and calculus.

We are given the function $f(x)=\frac{x}{(1+x^n{)}^{1/n}}$ and asked to find the $n$-fold composition $g(x)=(f\circ f\circ ...\circ f)(x)$.

At first glance, the idea of composing a function $n$ times feels daunting. But in mathematics, whenever you see a process repeated, look for a pattern. Let us compute the first composition, $f(f(x))$. We substitute $f(x)$ into itself:

Substituting the expression for $f(x)$ into the denominator, the fractional power $1/n$ cancels out with the exponent $n$, leaving us with $1+x^n$ in the denominator of the inner fraction.

When we simplify the expression $1+\frac{x^n}{1+x^n}$, we get $\frac{1+2x^n}{1+x^n}$. The $1+x^n$ terms cancel out, and we are left with $f(f(x))=\frac{x}{(1+2x^n{)}^{1/n}}$.

Do you see the magic? The coefficient of $x^n$ in the denominator has become $2$. If you were to compute $f(f(f(x)))$, you would find the coefficient becomes $3$.

By induction, the $n$-fold composition is simply:

We have tamed the beast!

Now that we have our function $g(x)$, we turn our attention to the integral: $I=\int x^{n-2}g(x)dx$. Substituting our newly found $g(x)$, we get:

Look closely at this integral. We have a term $x^{n-1}$ in the numerator and an expression involving $x^n$ in the denominator.

This is the hallmark of a perfect substitution. The derivative of $x^n$ is $n{x}^{n-1}$, which is exactly what we have in the numerator (up to a constant factor).

Let us set $u=1+n{x}^n$. Differentiating with respect to $x$, we get $du=n^2{x}^{n-1}dx$.

Rearranging this, we find $x^{n-1}dx=\frac{du}{n^2}$. Now, our integral becomes much friendlier:

This is a standard power rule integration. We add $1$ to the exponent and divide by the new exponent:

Simplifying the constants, the $n$ in the denominator of the fraction cancels one $n$ in the $n^2$, leaving us with $\frac{1}{n(n-1)}$.

Finally, we back-substitute $u=1+n{x}^n$ to arrive at our destination:

And there it is! A complex problem reduced to a simple, elegant result. Remember, in JEE, the goal isn't just to solve; it's to see the underlying structure.

Keep practicing, keep visualizing, and most importantly, keep falling in love with the process! The final result is $I=\frac{1}{n(n-1)}(1+n{x}^n{)}^{\frac{n-1}{n}}+K$.