202211041001 Solution to 2007-CE-AMATH-II-1

Find $\displaystyle{\int\frac{x^4+1}{x^2}\,\mathrm{d}x}$ .

Roughwork.

$\begin{aligned} &\quad\, \int\frac{x^4+1}{x^2}\,\mathrm{d}x\\ & = \int (x^2+x^{-2})\,\mathrm{d}x \\ & = \frac{x^3}{3} + \frac{x^{-1}}{(-1)} + C\textrm{ for some constant} \\ & = \frac{x^3}{3} - \frac{1}{x} + C \\ \end{aligned}$

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