計算不定積分 ∫ 1/(x+x^9) dx?
回答 (2)
2016-06-20 3:42 pm
∫ 1/(x + x⁹) dx
= ∫ 1/[x(1 + x⁸)] dx
= ∫ [1/x - x⁷/(1 + x⁸)] dx
= ∫ 1/x dx - ∫ x⁷/(1 + x⁸) dx
= ln|x| - (1/8) ∫ 1/(1 + x⁸) d(1 + x⁸)
= ln|x| - (1/8) ln|1 + x⁸| + C
= ln|x| - (1/8) ln(1 + x⁸) + C
= ∫ 1/[x(1 + x⁸)] dx
= ∫ [1/x - x⁷/(1 + x⁸)] dx
= ∫ 1/x dx - ∫ x⁷/(1 + x⁸) dx
= ln|x| - (1/8) ∫ 1/(1 + x⁸) d(1 + x⁸)
= ln|x| - (1/8) ln|1 + x⁸| + C
= ln|x| - (1/8) ln(1 + x⁸) + C
2016-06-21 2:50 am
∫ 1/(x+x^9) dx
=∫ (1+9x^8)/(x+x^9) d(x+x^9)
=In|x+x^9| +C
=∫ (1+9x^8)/(x+x^9) d(x+x^9)
=In|x+x^9| +C
收錄日期: 2021-04-18 15:08:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160620071625AAvJu85