Whakaoti i te ∫ 5(x+2)-(x-1)(x+4)-6x

$$\int 5 ( x + 2 ) - ( x - 1 ) ( x + 4 ) - 6 x$$

Aromātai

$$-\frac{x^{3}}{3}-2x^{2}+14x+С$$

Ngā Upane Otinga

Whakamahia te āhuatanga tohatoha hei whakarea te 5 ki te x+2.

$$\int 5x+10-\left(x-1\right)\left(x+4\right)-6x\mathrm{d}x$$

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o x-1 ki ia tau o x+4.

$$\int 5x+10-\left(x^{2}+4x-x-4\right)-6x\mathrm{d}x$$

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

$$\int 5x+10-\left(x^{2}+3x-4\right)-6x\mathrm{d}x$$

Hei kimi i te tauaro o x^{2}+3x-4, kimihia te tauaro o ia taurangi.

$$\int 5x+10-x^{2}-3x-\left(-4\right)-6x\mathrm{d}x$$

Ko te tauaro o -4 ko 4.

$$\int 5x+10-x^{2}-3x+4-6x\mathrm{d}x$$

Pahekotia te 5x me -3x, ka 2x.

$$\int 2x+10-x^{2}+4-6x\mathrm{d}x$$

Tāpirihia te 10 ki te 4, ka 14.

$$\int 2x+14-x^{2}-6x\mathrm{d}x$$

Pahekotia te 2x me -6x, ka -4x.

$$\int -4x+14-x^{2}\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int -4x\mathrm{d}x+\int 14\mathrm{d}x+\int -x^{2}\mathrm{d}x$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$-4\int x\mathrm{d}x+\int 14\mathrm{d}x-\int x^{2}\mathrm{d}x$$

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 -4 ki te \frac{x^{2}}{2}.

$$-2x^{2}+\int 14\mathrm{d}x-\int x^{2}\mathrm{d}x$$

Kimihia te tau tōpū o 14 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

$$-2x^{2}+14x-\int x^{2}\mathrm{d}x$$

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 -1 ki te \frac{x^{3}}{3}.

$$-2x^{2}+14x-\frac{x^{3}}{3}$$

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.

$$-2x^{2}+14x-\frac{x^{3}}{3}+С$$

Kimi Pārōnaki e ai ki x

$$14-4x-x^{2}$$