Whakaoti i te ∫ (x^3-2x^2+1/sqrt[3]{x^2}) wrt x

$$\int ( x ^ { 3 } - 2 x ^ { 2 } + \frac { 1 } { \sqrt[ 3 ] { x ^ { 2 } } } ) d x$$

Aromātai

$$\frac{x^{4}}{4}-\frac{2x^{3}}{3}+3\sqrt[3]{x}+С$$

Ngā Upane Otinga

Kōmitimititia te kīanga tapeke mā te kīanga.

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

Whakatauwehea te pūmau i ēnei kīanga katoa.

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

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}.

$$\frac{x^{4}}{4}-2\int x^{2}\mathrm{d}x+\int \frac{1}{x^{\frac{2}{3}}}\mathrm{d}x$$

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 -2 ki te \frac{x^{3}}{3}.

$$\frac{x^{4}}{4}-\frac{2x^{3}}{3}+\int \frac{1}{x^{\frac{2}{3}}}\mathrm{d}x$$

Tuhia anō te \frac{1}{x^{\frac{2}{3}}} hei x^{-\frac{2}{3}}. Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int x^{-\frac{2}{3}}\mathrm{d}x ki te \frac{x^{\frac{1}{3}}}{\frac{1}{3}}. Whakarūnāhia me te tahuri mai i te āhua taupū ki te āhua pūtake.

$$\frac{x^{4}}{4}-\frac{2x^{3}}{3}+3\sqrt[3]{x}$$

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.

$$\frac{x^{4}}{4}-\frac{2x^{3}}{3}+3\sqrt[3]{x}+С$$

Kimi Pārōnaki e ai ki x

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