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\int \left(x^{2}\right)^{3}+6\left(x^{2}\right)^{2}+12x^{2}+8\mathrm{d}x
\left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} binom teoremasini \left(x^{2}+2\right)^{3} kengaytirilishi uchun ishlating.
\int x^{6}+6\left(x^{2}\right)^{2}+12x^{2}+8\mathrm{d}x
Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 3 ni ko‘paytirib, 6 ni oling.
\int x^{6}+6x^{4}+12x^{2}+8\mathrm{d}x
Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 2 ni ko‘paytirib, 4 ni oling.
\int x^{6}\mathrm{d}x+\int 6x^{4}\mathrm{d}x+\int 12x^{2}\mathrm{d}x+\int 8\mathrm{d}x
Summani muddatma-muddat integratsiya qiling.
\int x^{6}\mathrm{d}x+6\int x^{4}\mathrm{d}x+12\int x^{2}\mathrm{d}x+\int 8\mathrm{d}x
Har bir shartda konstantani qavsdan tashqariga oling.
\frac{x^{7}}{7}+6\int x^{4}\mathrm{d}x+12\int x^{2}\mathrm{d}x+\int 8\mathrm{d}x
k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x^{6}\mathrm{d}x integralni \frac{x^{7}}{7} bilan almashtiring.
\frac{x^{7}}{7}+\frac{6x^{5}}{5}+12\int x^{2}\mathrm{d}x+\int 8\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. 6 ni \frac{x^{5}}{5} marotabaga ko'paytirish.
\frac{x^{7}}{7}+\frac{6x^{5}}{5}+4x^{3}+\int 8\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. 12 ni \frac{x^{3}}{3} marotabaga ko'paytirish.
\frac{x^{7}}{7}+\frac{6x^{5}}{5}+4x^{3}+8x
\int a\mathrm{d}x=ax umumiy integrallar qoidasi jadvalidan foydalanib, 8 integralini toping.
8x+4x^{3}+\frac{6x^{5}}{5}+\frac{x^{7}}{7}
Qisqartirish.
8x+4x^{3}+\frac{6x^{5}}{5}+\frac{x^{7}}{7}+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.