∫ (x+sqrt[3]{x}+1/x^2) wrt x eching

$$\int ( x + \sqrt[ 3 ] { x } + \frac { 1 } { x ^ { 2 } } ) d x$$

Baholash

$$\frac{x^{2}}{2}+\frac{3x^{\frac{4}{3}}}{4}-\frac{1}{x}+С$$

Yechim qadamlari

Summani muddatma-muddat integratsiya qiling.

$$\int x\mathrm{d}x+\int \sqrt[3]{x}\mathrm{d}x+\int \frac{1}{x^{2}}\mathrm{d}x$$

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.

$$\frac{x^{2}}{2}+\int \sqrt[3]{x}\mathrm{d}x+\int \frac{1}{x^{2}}\mathrm{d}x$$

\sqrt[3]{x} ni x^{\frac{1}{3}} sifatida qaytadan yozish. k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x^{\frac{1}{3}}\mathrm{d}x integralni \frac{x^{\frac{4}{3}}}{\frac{4}{3}} bilan almashtiring. Qisqartirish.

$$\frac{x^{2}}{2}+\frac{3x^{\frac{4}{3}}}{4}+\int \frac{1}{x^{2}}\mathrm{d}x$$

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

$$\frac{x^{2}}{2}+\frac{3x^{\frac{4}{3}}}{4}-\frac{1}{x}$$

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.

$$\frac{x^{2}}{2}+\frac{3x^{\frac{4}{3}}}{4}-\frac{1}{x}+С$$

x ga nisbatan hosilani topish

$$x+\sqrt[3]{x}+\frac{1}{x^{2}}$$