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Aromātai
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\int x^{2}+\frac{1}{x^{2}}\mathrm{d}x
Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.
\int x^{2}\mathrm{d}x+\int \frac{1}{x^{2}}\mathrm{d}x
Kōmitimititia te kīanga tapeke mā te kīanga.
\frac{x^{3}}{3}+\int \frac{1}{x^{2}}\mathrm{d}x
Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int x^{2}\mathrm{d}x ki te \frac{x^{3}}{3}.
\frac{x^{3}}{3}-\frac{1}{x}
Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int \frac{1}{x^{2}}\mathrm{d}x ki te -\frac{1}{x}.
\frac{4^{3}}{3}-4^{-1}-\left(\frac{1^{3}}{3}-1^{-1}\right)
Ko te tau tōpū tautuhi ko te pārōnaki kōaro o te kīanga i aromātaitia i te tepe tōrunga o te pāwhaitua, tangohia te pārōnaki kōaro i aromātaitia i te tepe tōraro o te pāwhaitua.
\frac{87}{4}
Whakarūnātia.