Whakaoti i te ∫ x^2-7x+12/x-3 wrt x=

$$\int \frac { x ^ { 2 } - 7 x + 12 } { x - 3 } d x =$$

Aromātai

$$\frac{x^{2}}{2}-4x+С$$

Ngā Upane Otinga

Me whakatauwehe ngā kīanga kāore anō i whakatauwehea i roto o \frac{x^{2}-7x+12}{x-3}.

$$\int \frac{\left(x-4\right)\left(x-3\right)}{x-3}\mathrm{d}x$$

Me whakakore tahi te x-3 i te taurunga me te tauraro.

$$\int x-4\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int x\mathrm{d}x+\int -4\mathrm{d}x$$

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}.

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

Kimihia te tau tōpū o -4 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

$$\frac{x^{2}}{2}-4x$$

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.

$$-4x+\frac{x^{2}}{2}+С$$

Kimi Pārōnaki e ai ki x

$$x-4$$