Integrate \( \frac{1+x}{x^3}\,\mathrm{d}x \)

∫1+xx3dx $\int \frac{1+x}{x^3}\mathrm{d}x$

= ∫(1x3+xx3)dx $\int (\frac{1}{x^3}+\frac{x}{x^3})\mathrm{d}x$

= ∫(x−3+x−2)dx $\int (x^{-3}+x^{-2})\mathrm{d}x$

= −12x2−1x+k