∫ x(2+x^2)^2 wrt x eching

$$\int x ( 2 + x ^ { 2 } ) ^ { 2 } d x$$

Baholash

$$\frac{x^{6}}{6}+x^{4}+2x^{2}+С$$

Yechim qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(2+x^{2}\right)^{2} kengaytirilishi uchun ishlating.

$$\int x\left(4+4x^{2}+\left(x^{2}\right)^{2}\right)\mathrm{d}x$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 2 ni ko‘paytirib, 4 ni oling.

$$\int x\left(4+4x^{2}+x^{4}\right)\mathrm{d}x$$

x ga 4+4x^{2}+x^{4} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\int 4x+4x^{3}+x^{5}\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int 4x\mathrm{d}x+\int 4x^{3}\mathrm{d}x+\int x^{5}\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$4\int x\mathrm{d}x+4\int x^{3}\mathrm{d}x+\int x^{5}\mathrm{d}x$$

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.

$$2x^{2}+4\int x^{3}\mathrm{d}x+\int x^{5}\mathrm{d}x$$

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x^{3}\mathrm{d}x integralni \frac{x^{4}}{4} bilan almashtiring. 4 ni \frac{x^{4}}{4} marotabaga ko'paytirish.

$$2x^{2}+x^{4}+\int x^{5}\mathrm{d}x$$

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x^{5}\mathrm{d}x integralni \frac{x^{6}}{6} bilan almashtiring.

$$2x^{2}+x^{4}+\frac{x^{6}}{6}$$

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.

$$\frac{x^{6}}{6}+x^{4}+2x^{2}+С$$

x ga nisbatan hosilani topish

$$x\left(x^{2}+2\right)^{2}$$