∫ (x+8)^2 wrt x eching

$$\int ( x + 8 ) ^ { 2 } d x$$

Baholash

$$\frac{\left(x+8\right)^{3}}{3}+С$$

Yechim qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+8\right)^{2} kengaytirilishi uchun ishlating.

$$\int x^{2}+16x+64\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int x^{2}\mathrm{d}x+\int 16x\mathrm{d}x+\int 64\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$\int x^{2}\mathrm{d}x+16\int x\mathrm{d}x+\int 64\mathrm{d}x$$

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x^{2}\mathrm{d}x integralni \frac{x^{3}}{3} bilan almashtiring.

$$\frac{x^{3}}{3}+16\int x\mathrm{d}x+\int 64\mathrm{d}x$$

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

$$\frac{x^{3}}{3}+8x^{2}+\int 64\mathrm{d}x$$

\int a\mathrm{d}x=ax umumiy integrallar qoidasi jadvalidan foydalanib, 64 integralini toping.

$$\frac{x^{3}}{3}+8x^{2}+64x$$

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.

$$\frac{x^{3}}{3}+8x^{2}+64x+С$$

x ga nisbatan hosilani topish

$$\left(x+8\right)^{2}$$