∫ intxe^xz wrt x eching

$$\int \int x e ^ { x z } d x$$

Baholash

$$\left\{\begin{matrix}\frac{xze^{xz}-2e^{xz}}{z^{3}}+С\left(x+1\right),&z\neq 0\\Сx+\frac{x^{3}}{6}+С_{1},&z=0\end{matrix}\right,$$

Yechim qadamlari

Qisqartirish.

$$\left(\frac{-e^{xz}+xze^{xz}}{z^{2}}+С_{3}\right)x-\frac{x^{2}e^{xz}}{z}+\frac{2\left(-e^{xz}+xze^{xz}\right)}{z^{3}}$$

Summani muddatma-muddat integratsiya qiling.

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

Har bir shartda konstantani qavsdan tashqariga oling.

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

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{1}{2} ni \frac{x^{3}}{3} marotabaga ko'paytirish.

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

\int a\mathrm{d}x=ax umumiy integrallar qoidasi jadvalidan foydalanib, С_{4} integralini toping.

$$\frac{x^{3}}{6}+С_{4}x$$

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.

$$\left\{\begin{matrix}\left(\frac{-e^{xz}+xze^{xz}}{z^{2}}+С_{3}\right)x-\frac{x^{2}e^{xz}}{z}+\frac{2\left(-e^{xz}+xze^{xz}\right)}{z^{3}}+С_{7},&\\\frac{x^{3}}{6}+С_{4}x+С_{7},&\end{matrix}\right,$$

x ga nisbatan hosilani topish

$$\left\{\begin{matrix}\frac{xze^{xz}-e^{xz}}{z^{2}}+С,&z\neq 0\\\frac{x^{2}}{2}+С,&z=0\end{matrix}\right,$$