Whakaoti i te ∫ x^4+x^2+1/x^2+x+1 wrt x

$$\int \frac { x ^ { 4 } + x ^ { 2 } + 1 } { x ^ { 2 } + x + 1 } d x$$

Aromātai

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

Ngā Upane Otinga

Me whakatauwehe ngā kīanga kāore anō i whakatauwehea i roto o \frac{x^{4}+x^{2}+1}{x^{2}+x+1}.

$$\int \frac{\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)}{x^{2}+x+1}\mathrm{d}x$$

Me whakakore tahi te x^{2}+x+1 i te taurunga me te tauraro.

$$\int x^{2}-x+1\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int x^{2}\mathrm{d}x+\int -x\mathrm{d}x+\int 1\mathrm{d}x$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$\int x^{2}\mathrm{d}x-\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}-\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}-\frac{x^{2}}{2}+\int 1\mathrm{d}x$$

Kimihia te tau tōpū o 1 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

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

Whakarūnātia.

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

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.

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

Kimi Pārōnaki e ai ki x

$$x^{2}-x+1$$