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\int \frac{x\left(x-2\right)\left(x+2\right)\left(x^{2}+5\right)}{x+2}\mathrm{d}x
\frac{x^{5}+x^{3}-20x}{x+2} ichida hali faktorlanmagan ifodalarni faktorlang.
\int x\left(x-2\right)\left(x^{2}+5\right)\mathrm{d}x
Surat va maxrajdagi ikkala x+2 ni qisqartiring.
\int x^{4}-2x^{3}+5x^{2}-10x\mathrm{d}x
Ifodani kengaytiring.
\int x^{4}\mathrm{d}x+\int -2x^{3}\mathrm{d}x+\int 5x^{2}\mathrm{d}x+\int -10x\mathrm{d}x
Summani muddatma-muddat integratsiya qiling.
\int x^{4}\mathrm{d}x-2\int x^{3}\mathrm{d}x+5\int x^{2}\mathrm{d}x-10\int x\mathrm{d}x
Har bir shartda konstantani qavsdan tashqariga oling.
\frac{x^{5}}{5}-2\int x^{3}\mathrm{d}x+5\int x^{2}\mathrm{d}x-10\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^{4}\mathrm{d}x integralni \frac{x^{5}}{5} bilan almashtiring.
\frac{x^{5}}{5}-\frac{x^{4}}{2}+5\int x^{2}\mathrm{d}x-10\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. -2 ni \frac{x^{4}}{4} marotabaga ko'paytirish.
\frac{x^{5}}{5}-\frac{x^{4}}{2}+\frac{5x^{3}}{3}-10\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^{2}\mathrm{d}x integralni \frac{x^{3}}{3} bilan almashtiring. 5 ni \frac{x^{3}}{3} marotabaga ko'paytirish.
\frac{x^{5}}{5}-\frac{x^{4}}{2}+\frac{5x^{3}}{3}-5x^{2}
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. -10 ni \frac{x^{2}}{2} marotabaga ko'paytirish.
-5x^{2}+\frac{5x^{3}}{3}-\frac{x^{4}}{2}+\frac{x^{5}}{5}
Qisqartirish.
-5x^{2}+\frac{5x^{3}}{3}-\frac{x^{4}}{2}+\frac{x^{5}}{5}+С
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.