Whakaoti i te ∫ x^2(x+1)^3 wrt x

$$\int x ^ { 2 } ( x + 1 ) ^ { 3 } d x$$

Aromātai

$$\frac{x^{6}}{6}+\frac{3x^{5}}{5}+\frac{3x^{4}}{4}+\frac{x^{3}}{3}+С$$

Ngā Upane Otinga

Whakamahia te ture huarua \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} hei whakaroha \left(x+1\right)^{3}.

$$\int x^{2}\left(x^{3}+3x^{2}+3x+1\right)\mathrm{d}x$$

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2} ki te x^{3}+3x^{2}+3x+1.

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

Kōmitimititia te kīanga tapeke mā te kīanga.

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

Whakatauwehea te pūmau i ēnei kīanga katoa.

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

Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int x^{5}\mathrm{d}x ki te \frac{x^{6}}{6}.

$$\frac{x^{6}}{6}+3\int x^{4}\mathrm{d}x+3\int x^{3}\mathrm{d}x+\int x^{2}\mathrm{d}x$$

Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int x^{4}\mathrm{d}x ki te \frac{x^{5}}{5}. Whakareatia 3 ki te \frac{x^{5}}{5}.

$$\frac{x^{6}}{6}+\frac{3x^{5}}{5}+3\int x^{3}\mathrm{d}x+\int x^{2}\mathrm{d}x$$

Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int x^{3}\mathrm{d}x ki te \frac{x^{4}}{4}. Whakareatia 3 ki te \frac{x^{4}}{4}.

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

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}.

$$\frac{x^{6}}{6}+\frac{3x^{5}}{5}+\frac{3x^{4}}{4}+\frac{x^{3}}{3}$$

Whakarūnātia.

$$\frac{x^{3}}{3}+\frac{3x^{4}}{4}+\frac{3x^{5}}{5}+\frac{x^{6}}{6}$$

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.

$$\frac{x^{3}}{3}+\frac{3x^{4}}{4}+\frac{3x^{5}}{5}+\frac{x^{6}}{6}+С$$

Kimi Pārōnaki e ai ki x

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