JAMB 2013 · UTME · Q38

Integrate $\dfrac{1 + x}{x^3}\,dx$ .

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 $-\frac{1}{2x^2} - \frac1x + k$ , option B.