Question:

Step 1: Understand the function $f(x)$ and its composition $g(x)$. The given function is $f(x) = \frac{x}{(1 + x^n)^{1/n}}$, where $n \ge 2$. The function $g(x)$ is defined as the $n$-th composition of $f(x)$, i.e., $g(x) = (f \circ f \circ \dots \circ f)(x)$ where $f$ occurs $n$ times. We need to evaluate the integral $\int x^{n-2} g(x) dx$. Step 2: Calculate the composition $f(f(x))$. To find a pattern for $g(x)$, we first compute $f(f(x))$. We substitute $f(x)$ into $f(x)$: $$f(f(x)) = \frac{f(x)}{(1 + (f(x))^n)^{1/n}}$$ Now substitute the expression for $f(x) = \frac{x}{(1 + x^n)^{1/n}}$: $$f(f(x)) = \frac{\frac{x}{(1 + x^n)^{1/n}}}{\left(1 + \left(\frac{x}{(1 + x^n)^{1/n}}\right)^n\right)^{1/n}}$$ First, simplify the term $(f(x))^n$: $$(f(x))^n = \left(\frac{x}{(1 + x^n)^{1/n}}\right)^n = \frac{x^n}{((1 + x^n)^{1/n})^n} = \frac{x^n}{1 + x^n}$$ Substitute this back into the expression for $f(f(x))$: $$f(f(x)) = \frac{\frac{x}{(1 + x^n)^{1/n}}}{\left(1 + \frac{x^n}{1 + x^n}\right)^{1/n}}$$ Now, simplify the expression inside the parenthesis in the denominator: $$1 + \frac{x^n}{1 + x^n} = \frac{(1 + x^n) + x^n}{1 + x^n} = \frac{1 + 2x^n}{1 + x^n}$$ Substitute this back: $$f(f(x)) = \frac{\frac{x}{(1 + x^n)^{1/n}}}{\left(\frac{1 + 2x^n}{1 + x^n}\right)^{1/n}}$$ $$f(f(x)) = \frac{x}{(1 + x^n)^{1/n}} \cdot \frac{(1 + x^n)^{1/n}}{(1 + 2x^n)^{1/n}}$$ $$f(f(x)) = \frac{x}{(1 + 2x^n)^{1/n}}$$ Step 3: Generalize $g(x)$ using the observed pattern. We have found the following pattern for repeated compositions: $f(x) = \frac{x}{(1 + 1 \cdot x^n)^{1/n}}$ $f(f(x)) = \frac{x}{(1 + 2 \cdot x^n)^{1/n}}$ By induction, applying the function $f$ 'k' times yields the general formula: $$(f \circ f \circ \dots \circ f)(x) \text{ (k times)} = \frac{x}{(1 + k \cdot x^n)^{1/n}}$$ Since $g(x) = (f \circ f \circ \dots \circ f)(x)$ where $f$ occurs $n$ times, we set $k=n$: $$g(x) = \frac{x}{(1 + nx^n)^{1/n}}$$ Step 4: Substitute $g(x)$ into the integral and simplify the integrand. The integral we need to evaluate is $I = \int x^{n-2} g(x) dx$. Substitute the derived expression for $g(x)$: $$I = \int x^{n-2} \cdot \frac{x}{(1 + nx^n)^{1/n}} dx$$ Combine the powers of $x$ in the numerator: $$I = \int \frac{x^{n-1}}{(1 + nx^n)^{1/n}} dx$$ This can be written as: $$I = \int x^{n-1} (1 + nx^n)^{-1/n} dx$$ Step 5: Evaluate the integral using a suitable substitution. To solve this integral, we use the substitution method. Let $u = 1 + nx^n$. Now, we find the differential $du$: $$\frac{du}{dx} = \frac{d}{dx}(1 + nx^n) = 0 + n \cdot (nx^{n-1}) = n^2 x^{n-1}$$ So, $du = n^2 x^{n-1} dx$. From this, we can express $x^{n-1} dx$ in terms of $du$: $$x^{n-1} dx = \frac{1}{n^2} du$$ Substitute $u$ and $x^{n-1} dx$ into the integral: $$I = \int (u)^{-1/n} \left(\frac{1}{n^2} du\right)$$ $$I = \frac{1}{n^2} \int u^{-1/n} du$$ Now, integrate $u^{-1/n}$ with respect to $u$: $$I = \frac{1}{n^2} \cdot \frac{u^{-1/n + 1}}{-1/n + 1} + K$$ Simplify the exponent and the denominator: $$-1/n + 1 = \frac{-1 + n}{n} = \frac{n-1}{n}$$ So the integral becomes: $$I = \frac{1}{n^2} \cdot \frac{u^{(n-1)/n}}{(n-1)/n} + K$$ $$I = \frac{1}{n^2} \cdot \frac{n}{n-1} u^{(n-1)/n} + K$$ $$I = \frac{1}{n(n-1)} u^{(n-1)/n} + K$$ Finally, substitute back $u = 1 + nx^n$: $$I = \frac{1}{n(n-1)} (1 + nx^n)^{(n-1)/n} + K$$ The exponent $(n-1)/n$ can also be written as $1 - 1/n$: $$I = \frac{1}{n(n-1)} (1 + nx^n)^{1 - 1/n} + K$$ Step 6: State the final answer and match it with the correct option. The evaluated integral is $\frac{1}{n(n-1)} (1 + nx^n)^{1 - 1/n} + K$. This matches Option 1 (A). The final answer is $\boxed{\text{1/(n(n-1)) * (1 + nx^n)^(1 - 1/n) + K}}$.