这个积分怎么算∫（1-x^2）/（x(1+x^2)）dx=?

∫(1-x^2)/[x(1+x^2)]dxlet(1-x^2)/[x(1+x^2)] = a/x+ (b1x+b0)/(1+x^2)1-x^2= a(1+x^2) +x(b1x+b0)put x=0a=1coef。of x^2-1=a+b1b1=-2coef。of xb0=0(1-x^2)/[x(1+x^2)] = 1/x- 2x/(1+x^2)∫(1-x^2)/[x(1+x^2)]dx=∫[1/x- 2x/(1+x^2)] dx= ln|x| - ln(1+x^2) + C