1/(1+2x)(1+x^2)的不定积分

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/30 19:41:06
1/(1+2x)(1+x^2)的不定积分
x[kAǿ(&$s^% QTvv7jوPAl"^ZPP("D]4ԧ~䢭݇sgv*qXz:( ߟgGl»Q*ҝ|~Kb.WQ om=wCKDn/@n(. :pYC\i(RA3O$1:6%ZEEd,)l*O^e㧢=p{ɊHcr ኞٰ{.SBD(?_=V{GII*+ښS Nb]߅L6čC6z.bbjyE 6+ nD!2S҂N}hz2;8zz0wQsX7ͯMoyIRE7oPV_8,uvl㻭(:Nk|"d5Bpm~:e؜NЦM&Pm.10XmpV`VF V y1LZzac9Žr EN!ouh

1/(1+2x)(1+x^2)的不定积分
1/(1+2x)(1+x^2)的不定积分

1/(1+2x)(1+x^2)的不定积分
令1/[(1 + 2x)(1 + x^2)] = A/(1 + 2x) + (Bx + C)/(1 + x^2)
则1 = A(1 + x^2) + (Bx + C)(1 + 2x)
1 = (A + 2B)x^2 + (B + 2C)x + (A + C)
A + 2B = 0、B + 2C = 0、A + C = 1
解得A = 4/5、B = - 2/5、C = 1/5
原式 = (4/5)∫ dx/(1 + 2x) - (2/5)∫ x/(1 + x^2) dx + (1/5)∫ dx/(1 + x^2)
= (4/5)(1/2)∫ d(1 + 2x)/(1 + 2x) - (2/5)(1/2)∫ d(1 + x^2)/(1 + x^2) + (1/5)∫ dx/(1 + x^2)
= (2/5)ln|1 + 2x| - (1/5)ln(1 + x^2) + (1/5)arctan(x) + C

S(1/(1+2x)(1+x^2))dx
=S(4/5*1/(1+2x)-1/5*(2x-1)/(x^2+1))dx
=2/5*ln(1+2x)-1/5*ln(x^2+1)-1/5*arctanx+c
希望对你有所帮助 还望采纳~~~