Question

求不定积分∫(arctanx)/(x^2(x^2+1))dx

Answer 1

求不定积分∫{(arctanx)/[x²(x²+1)]}dx

解：原式=∫[(arctanx)/x²-(arctanx)/(1+x²)]dx=∫[(arctanx)/x²]dx-∫[(arctanx)/(1+x²)]dx
=-∫(arctanx)d(1/x)-∫(arctanx)d(arctanx)=-{(1/x)arctanx-∫dx/[x(1+x²)]}-(1/2)(arctanx)²
=-(1/x)arctanx+∫[(1/x)-x/(x²+1)]dx-(1/2)(arctanx)²
=-(1/x)arctanx+∫(1/x)dx-(1/2)∫d(x²+1)/(x²+1)-(1/2)(arctanx)²
=-(1/x)arctanx+ln∣x∣-(1/2)ln(x²+1)-(1/2)(arctanx)²+C

Answer 2

∫(arctanx)/(x^2(x^2+1))dx

=∫(arctanx)/x^2dx-∫(arctanx)/(x^2+1)dx
=∫(arctanx)d(1/x)-∫(arctanx)darctanx
=arctanx/x-∫1/xdarctanx-1/2(arctanx)^2
=arctanx/x-1/2(arctanx)^2-∫1/[x(x^2+1)]dx
=arctanx/x-1/2(arctanx)^2-∫[1/x-x/(x^2+1)]dx
=arctanx/x-1/2(arctanx)^2-lnx+1/2ln(x^2+1)+C