∫ x^5+x^3-20x eching

$$\int x ^ { 5 } + x ^ { 3 } - 20 x$$

Baholash

$$\frac{x^{6}}{6}+\frac{x^{4}}{4}-10x^{2}+С$$

Yechim qadamlari

Summani muddatma-muddat integratsiya qiling.

$$\int x^{5}\mathrm{d}x+\int x^{3}\mathrm{d}x+\int -20x\mathrm{d}x$$

Har bir shartda konstantani qavsdan tashqariga oling.

$$\int x^{5}\mathrm{d}x+\int x^{3}\mathrm{d}x-20\int x\mathrm{d}x$$

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.

$$\frac{x^{6}}{6}+\int x^{3}\mathrm{d}x-20\int x\mathrm{d}x$$

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.

$$\frac{x^{6}}{6}+\frac{x^{4}}{4}-20\int x\mathrm{d}x$$

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. -20 ni \frac{x^{2}}{2} marotabaga ko'paytirish.

$$\frac{x^{6}}{6}+\frac{x^{4}}{4}-10x^{2}$$

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.

$$\frac{x^{6}}{6}+\frac{x^{4}}{4}-10x^{2}+С$$

x ga nisbatan hosilani topish

$$x\left(x^{4}+x^{2}-20\right)$$