∫ 2xleft(x^2+1right)^3 wrt x eching

$$\int{ 2x { \left( { x }^{ 2 } +1 \right) }^{ 3 } }d x$$

Baholash

$$\frac{x^{8}}{4}+x^{6}+\frac{3x^{4}}{2}+x^{2}+С$$

Yechim qadamlari

\left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} binom teoremasini \left(x^{2}+1\right)^{3} kengaytirilishi uchun ishlating.

$$\int 2x\left(\left(x^{2}\right)^{3}+3\left(x^{2}\right)^{2}+3x^{2}+1\right)\mathrm{d}x$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 3 ni ko‘paytirib, 6 ni oling.

$$\int 2x\left(x^{6}+3\left(x^{2}\right)^{2}+3x^{2}+1\right)\mathrm{d}x$$

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

$$\int 2x\left(x^{6}+3x^{4}+3x^{2}+1\right)\mathrm{d}x$$

2x ga x^{6}+3x^{4}+3x^{2}+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\int 2x^{7}+6x^{5}+6x^{3}+2x\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int 2x^{7}\mathrm{d}x+\int 6x^{5}\mathrm{d}x+\int 6x^{3}\mathrm{d}x+\int 2x\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$2\int x^{7}\mathrm{d}x+6\int x^{5}\mathrm{d}x+6\int x^{3}\mathrm{d}x+2\int x\mathrm{d}x$$

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x^{7}\mathrm{d}x integralni \frac{x^{8}}{8} bilan almashtiring. 2 ni \frac{x^{8}}{8} marotabaga ko'paytirish.

$$\frac{x^{8}}{4}+6\int x^{5}\mathrm{d}x+6\int x^{3}\mathrm{d}x+2\int x\mathrm{d}x$$

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. 6 ni \frac{x^{6}}{6} marotabaga ko'paytirish.

$$\frac{x^{8}}{4}+x^{6}+6\int x^{3}\mathrm{d}x+2\int x\mathrm{d}x$$

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. 6 ni \frac{x^{4}}{4} marotabaga ko'paytirish.

$$\frac{x^{8}}{4}+x^{6}+\frac{3x^{4}}{2}+2\int x\mathrm{d}x$$

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. 2 ni \frac{x^{2}}{2} marotabaga ko'paytirish.

$$\frac{x^{8}}{4}+x^{6}+\frac{3x^{4}}{2}+x^{2}$$

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.

$$x^{2}+\frac{3x^{4}}{2}+x^{6}+\frac{x^{8}}{4}+С$$

x ga nisbatan hosilani topish

$$2x\left(x^{2}+1\right)^{3}$$