∫ sqrt{ax wrt x=frac{2xsqrt{ax}}{3}+c eching

$$\int \sqrt { a x } d x = \frac { 2 x \sqrt { a x } } { 3 } + c$$

c uchun yechish

$$c=С$$

$$\left(x\leq 0\text{ and }a<0\right)\text{ or }\left(x\geq 0\text{ and }a>0\right)$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglamaning ikkala tarafini 3 ga ko'paytirish.

$$3\int \sqrt{ax}\mathrm{d}x=2x\sqrt{ax}+3c$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$2x\sqrt{ax}+3c=3\int \sqrt{ax}\mathrm{d}x$$

Ikkala tarafdan 2x\sqrt{ax} ni ayirish.

$$3c=3\int \sqrt{ax}\mathrm{d}x-2x\sqrt{ax}$$

Tenglama standart shaklda.

$$3c=3x\sqrt{ax}-2x\sqrt{ax}+3С$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3c}{3}=\frac{x\sqrt{ax}+3С}{3}$$

3 ga bo'lish 3 ga ko'paytirishni bekor qiladi.

$$c=\frac{x\sqrt{ax}+3С}{3}$$

\sqrt{ax}x+3С ni 3 ga bo'lish.

$$c=\frac{x\sqrt{ax}}{3}+С$$

a uchun yechish

$$\left\{\begin{matrix}a=\frac{С_{1}^{2}+9c^{2}-Сc}{x^{3}}\text{, }&\left(c\leq С\text{ and }x<0\right)\text{ or }\left(c\geq С_{1}\text{ and }x>0\right)\\a\in \mathrm{R}\text{, }&x=0\text{ and }c=С\end{matrix}\right,$$