(A) Let $f(x) = t$. Then,$f'(x) \, dx = dt$. Substituting these into the integral,we get: $\int \frac{f'(x)}{[f(x)]^2} \, dx = \int \frac{1}{t^2} \, d

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(A) Let $f(x) = t$. Then,$f'(x) \, dx = dt$. Substituting these into the integral,we get: $\int \frac{f'(x)}{[f(x)]^2} \, dx = \int \frac{1}{t^2} \, dt$ $= \int t^{-2} \, dt$ $= \frac{t^{-2+1}}{-2+1} + c$ $= \frac{t^{-1}}{-1} + c$ $= -\frac{1}{t} + c$ $= -\frac{1}{f(x)} + c$ $= -[f(x)]^{-1} + c$.