用分部积分法：∫udv=uv-∫vdu,
所以
∫ln(a^2+x^2)dx
=xln(a^2+x^2)-∫xd[ln(a^2+x^2)]
=xln(a^2+x^2)-∫[(2x^2)/(a^2+x^2)]dx
=xln(a^2+x^2)-∫[(2x^2+2a^2-2a^2)/(a^2+x^2)]dx
=xln(a^2+x^2)-∫[2-(2a^2)/(a^2+x^2)]dx
=xln(a^2+x^2)-2x+(2a^2)∫[1/(a^2+x^2)]dx
=xln(a^2+x^2)-2x+(2a^2)[(1/a)arctan(x/a)+C']
=xln(a^2+x^2)-2x+(2a)[arctan(x/a)]+C