积分:2/(1-u^2+2u)du
=积分:2/[-(u^2-2u+1)+2]du
=-2积分:1/[(u-1)^2-(根号2)^2]du
=-2积分:1/[(u-1)^2-(根号2)^2]d(u-1)
=-2/ln2*ln|(u-1-根号2)/(u-1+根号2)|+C
(C为常数)
有公式:
积分:dx/(x^2-a^2)
=1/2aln|(x-a)/(x+a)|+C
提示:
将:1/(x^2-a^2)=1/2a*(1/(x-a)-1/(x+a))
∫2/(1-u^2+2u)du怎么做
2个回答
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