Whakaoti i te ∫ [x+1)^{2}-4x]^2-2a(2a+1)^2-(x^2-2x)^2=

$$\int [ x + 1 ) ^ { 2 } - 4 x ] ^ { 2 } - 2 a ( 2 a + 1 ) ^ { 2 } - ( x ^ { 2 } - 2 x ) ^ { 2 } =$$

Aromātai

$$\frac{\left(x+1\right)^{3}}{3}-2x^{2}+С$$

Ngā Upane Otinga

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+1\right)^{2}.

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

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

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

Kōmitimititia te kīanga tapeke mā te kīanga.

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

Whakatauwehea te pūmau i ēnei kīanga katoa.

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

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}.

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

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

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

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.

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

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.

$$\frac{x^{3}}{3}-x^{2}+x+С$$

Kimi Pārōnaki e ai ki x

$$\left(x-1\right)^{2}$$