∫ (2sqrt{x-1})^2 wrt x eching

$$\int ( 2 \sqrt { x - 1 } ) ^ { 2 } d x$$

Baholash

$$2x^{2}-4x+С$$

Yechim qadamlari

\left(2\sqrt{x-1}\right)^{2} ni kengaytirish.

$$\int 2^{2}\left(\sqrt{x-1}\right)^{2}\mathrm{d}x$$

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

$$\int 4\left(\sqrt{x-1}\right)^{2}\mathrm{d}x$$

2 daraja ko‘rsatkichini \sqrt{x-1} ga hisoblang va x-1 ni qiymatni oling.

$$\int 4\left(x-1\right)\mathrm{d}x$$

4 ga x-1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\int 4x-4\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int 4x\mathrm{d}x+\int -4\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$4\int x\mathrm{d}x+\int -4\mathrm{d}x$$

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x\mathrm{d}x integralni \frac{x^{2}}{2} bilan almashtiring. 4 ni \frac{x^{2}}{2} marotabaga ko'paytirish.

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

\int a\mathrm{d}x=ax umumiy integrallar qoidasi jadvalidan foydalanib, -4 integralini toping.

$$2x^{2}-4x$$

Агар F\left(x\right)f\left(x\right) ning dastlabki holati boʻlsa, u holatda f\left(x\right) ning barcha dastlabki holatlari toʻplami F\left(x\right)+C tarafidan belgilanadi. Shu sababli natijaga C\in \mathrm{R} integrallash konstantasini qoʻshing.

$$2x^{2}-4x+С$$

x ga nisbatan hosilani topish

$$4\left(x-1\right)$$