Question

不定积分∫{1/[(x+a)^2*(x+b)^2]}dx是多少?

Answer 1

∫dx/[(x+a)(x+b)]^2

=(1/(a-b))∫[(x+a)-(x+b)]dx/[(x+a)(x+b)]^2

=(1/(a-b))∫dx/[(x+a)(x+b)^2] -(1/(a-b))∫dx/[(x+a)^2(x+b)]

=(1/(a-b)^2)∫dx/(x+b)^2-(1/(a-b)^2)∫dx/[(x+a)(x+b)]

-(1/(a-b)^2)∫dx/(x+a)(x+b)+(1/(a-b)^2)∫dx/(x+a)^2

=(1/(a-b)^2)(-1/(x+b)-1/(x+a) -2/(a-b)^3ln|(x+b)/(x+a)| +C