tajanaewilliams77
14.06.2021 •
Mathematics
Tính tích phân I=∫(x^2 + 1) ^3 dx
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Ответ:
Step-by-step explanation:
I=∫(x²+1)³dx
=∫[(x²)³+1³+3x²×1(x²+1)]dx
=∫(x^6+1+3x^4+3x²)dx
=∫(x^6+3x^4+3x²+1)dx
=(x^7)/7 +3(x^5)/5+3x³/3+x+c
Ответ: