反常积分求解

Question

如题所述

Answer 1

可以用分部积分先求不定积分
∫(arctanx)/x²dx=-∫arctanxd(1/x)
=-(arctanx)/x+∫1/xdarctanx
=-(arctanx)/x+∫1/[x(1+x²)]dx
=-(arctanx)/x+∫[1/x-x/(1+x²)]dx
=-(arctanx)/x+ln|x|-(1/2)∫1/(1+x²)dx²
=-(arctanx)/x+ln|x|-ln√(1+x²)+C
=-(arctanx)/x+ln[|x|/√(1+x²)]+C
∴其定积分=lim{x->+∞}[-(arctanx)/x+ln[|x|/√(1+x²)]]
-[-(arctan1)/1+ln(1/√2)]
=0+0-[-π/4-(1/2)ln2]
=π/4+(1/2)ln2