Whakaoti i te ∫ (x+3)(x-4) wrt x

$$\int ( x + 3 ) ( x - 4 ) d x$$

Aromātai

$$\frac{x^{3}}{3}-\frac{x^{2}}{2}-12x+С$$

Ngā Upane Otinga

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o x+3 ki ia tau o x-4.

$$\int x^{2}-4x+3x-12\mathrm{d}x$$

Pahekotia te -4x me 3x, ka -x.

$$\int x^{2}-x-12\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

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

Whakatauwehea te pūmau i ēnei kīanga katoa.

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

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}.

$$\frac{x^{3}}{3}-\int x\mathrm{d}x+\int -12\mathrm{d}x$$

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 -1 ki te \frac{x^{2}}{2}.

$$\frac{x^{3}}{3}-\frac{x^{2}}{2}+\int -12\mathrm{d}x$$

Kimihia te tau tōpū o -12 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}-\frac{x^{2}}{2}-12x$$

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.

$$\frac{x^{3}}{3}-\frac{x^{2}}{2}-12x+С$$

Kimi Pārōnaki e ai ki x

$$\left(x-4\right)\left(x+3\right)$$