Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 276: 87

$$\frac{-1}{f(x)}+C $$

Given $$\int f(x)^{3} f^{\prime}(x) \mathrm{d} x$$ Let $$ u=f(x) \ \ \ \Rightarrow \ \ \ du = f'(x) dx$$ Then \begin{aligned} \int \frac{f^{\prime}(x)}{f(x)^{2}} \mathrm{d} x &=\int \frac{d u}{u^{2}} \\ &=\frac{u^{-1}}{-1}+C \\ &=-\frac{1}{u}+C \\ &=\frac{-1}{f(x)}+C \end{aligned}