Whakaoti i te ∫ 12x^2+162x+623 wrt x

$$\int{ 12 { x }^{ 2 } +162x+623 }d x$$

Aromātai

$$4x^{3}+81x^{2}+623x+С$$

Ngā Upane Otinga

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int 12x^{2}\mathrm{d}x+\int 162x\mathrm{d}x+\int 623\mathrm{d}x$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$12\int x^{2}\mathrm{d}x+162\int x\mathrm{d}x+\int 623\mathrm{d}x$$

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}. Whakareatia 12 ki te \frac{x^{3}}{3}.

$$4x^{3}+162\int x\mathrm{d}x+\int 623\mathrm{d}x$$

Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int x\mathrm{d}x ki te \frac{x^{2}}{2}. Whakareatia 162 ki te \frac{x^{2}}{2}.

$$4x^{3}+81x^{2}+\int 623\mathrm{d}x$$

Kimihia te tau tōpū o 623 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

$$4x^{3}+81x^{2}+623x$$

Mēnā ko F\left(x\right) he pārōnaki kōaro o f\left(x\right), kāti ko te huinga o ngā pārōnaki kōaro katoa o f\left(x\right) ka whakaaturia e F\left(x\right)+C. Nō reira, me tāpiri te pūmau o te whakatōpūtanga C\in \mathrm{R} ki te otinga.

$$4x^{3}+81x^{2}+623x+С$$

Kimi Pārōnaki e ai ki x

$$12x^{2}+162x+623$$