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Aromātai
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Tohaina

\int \sqrt{1+\left(2t\right)^{2}}\mathrm{d}x
Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.
\sqrt{1+\left(2t\right)^{2}}x
Kimihia te tau tōpū o \sqrt{1+\left(2t\right)^{2}} mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.
\sqrt{1+4t^{2}}x
Whakarūnātia.
\left(1+4t^{2}\right)^{\frac{1}{2}}\times 1-\left(1+4t^{2}\right)^{\frac{1}{2}}\times 0
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.
\sqrt{1+4t^{2}}
Whakarūnātia.