∫ sqrt[3]{3t}dt=(3t)^4/2/4tc eching

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

c uchun yechish

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

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglamaning ikkala tarafini 4 ga ko'paytirish.

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

2 ni olish uchun 4 ni 2 ga bo‘ling.

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

\left(3t\right)^{2} ni kengaytirish.

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

2 daraja ko‘rsatkichini 3 ga hisoblang va 9 ni qiymatni oling.

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

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. 2 va 1 ni qo‘shib, 3 ni oling.

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

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

Tenglama standart shaklda.

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

Ikki tarafini 9t^{3} ga bo‘ling.

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

9t^{3} ga bo'lish 9t^{3} ga ko'paytirishni bekor qiladi.

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

\frac{4\times \left(3t\right)^{\frac{4}{3}}}{3}+4С ni 9t^{3} ga bo'lish.

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