Whakaoti i te ∫ (x-3)(x+1)

$$\int ( x - 3 ) ( x + 1 )$$

Aromātai

$$\frac{x^{3}}{3}-x^{2}-3x+С$$

Ngā Upane Otinga

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o x-3 ki ia tau o x+1.

$$\int x^{2}+x-3x-3\mathrm{d}x$$

Pahekotia te x me -3x, ka -2x.

$$\int x^{2}-2x-3\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int x^{2}\mathrm{d}x+\int -2x\mathrm{d}x+\int -3\mathrm{d}x$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$\int x^{2}\mathrm{d}x-2\int x\mathrm{d}x+\int -3\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}-2\int x\mathrm{d}x+\int -3\mathrm{d}x$$

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 -2 ki te \frac{x^{2}}{2}.

$$\frac{x^{3}}{3}-x^{2}+\int -3\mathrm{d}x$$

Kimihia te tau tōpū o -3 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}-x^{2}-3x$$

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.

$$\frac{x^{3}}{3}-x^{2}-3x+С$$

Kimi Pārōnaki e ai ki x

$$\left(x-3\right)\left(x+1\right)$$