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(x²-x-1)/(x²+1)的不定积分
人气:122 ℃ 时间:2020-05-14 17:38:58
解答
(x²-x-1)/(x²+1)
=(x²+1-x-2)/(x²+1)
=1-x/(x²+1)-2/(x²+1)
原式=∫1dx-∫x/(x²+1)-2∫1/(x²+1)dx
=∫1dx-1/2∫d(x²+1)/(x²+1)-2∫1/(x²+1)dx
=x-1/2*ln(x²+1)-2arctanx+C
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