Whakaoti i te ∫ sqrt[3]{3t}dt=(3t)^4/2/4tc

$$\int \sqrt[ 3 ] { 3 t } d t = \frac { ( 3 t ) ^ { 4 / 2 } } { 4 } t c$$

Whakaoti mō c

$$\left\{\begin{matrix}c=\frac{3^{\frac{4}{3}}}{9t^{\frac{5}{3}}}+\frac{4С}{9t^{3}}\text{, }&t\neq 0\\c\in \mathrm{R}\text{, }&С=0\text{ and }t=0\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakareatia ngā taha e rua o te whārite ki te 4.

$$4\int \sqrt[3]{3t}\mathrm{d}t=\left(3t\right)^{\frac{4}{2}}tc$$

Whakawehea te 4 ki te 2, kia riro ko 2.

$$4\int \sqrt[3]{3t}\mathrm{d}t=\left(3t\right)^{2}tc$$

Whakarohaina te \left(3t\right)^{2}.

$$4\int \sqrt[3]{3t}\mathrm{d}t=3^{2}t^{2}tc$$

Tātaihia te 3 mā te pū o 2, kia riro ko 9.

$$4\int \sqrt[3]{3t}\mathrm{d}t=9t^{2}tc$$

Hei whakarea i ngā pū o te pūtake kotahi, me tāpiri ō rātou taupū. Tāpiria te 2 me te 1 kia riro ai te 3.

$$4\int \sqrt[3]{3t}\mathrm{d}t=9t^{3}c$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$9t^{3}c=4\int \sqrt[3]{3t}\mathrm{d}t$$

He hanga arowhānui tō te whārite.

$$9t^{3}c=4\sqrt[3]{3}t^{\frac{4}{3}}+4С$$

Whakawehea ngā taha e rua ki te 9t^{3}.

$$\frac{9t^{3}c}{9t^{3}}=\frac{\frac{4\times \left(3t\right)^{\frac{4}{3}}}{3}+4С}{9t^{3}}$$

Mā te whakawehe ki te 9t^{3} ka wetekia te whakareanga ki te 9t^{3}.

$$c=\frac{\frac{4\times \left(3t\right)^{\frac{4}{3}}}{3}+4С}{9t^{3}}$$

Whakawehe \frac{4\times \left(3t\right)^{\frac{4}{3}}}{3}+4С ki te 9t^{3}.

$$c=\frac{4\left(\frac{\left(3t\right)^{\frac{4}{3}}}{3}+С\right)}{9t^{3}}$$