JAMB 1994 · UME · Q41

Integrate $\dfrac{1 - x}{x^3}$ with respect to $x$ .

Worked solution (try it first)

Split the fraction into powers of $x$ : $\dfrac{1 - x}{x^3} = x^{-3} - x^{-2}$ .
Add one to each power and divide by the new power: $x^{-3}$ gives $\frac{x^{-2}}{-2} = -\frac{1}{2x^2}$ , and $-x^{-2}$ gives $-\frac{x^{-1}}{-1} = \frac1x$ .
So the integral is $\frac1x - \frac{1}{2x^2} + k$ , option C.