求不定积分∫(arctanx)/(x^2(x^2+1))dx

网友回答

∫(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
======以下答案可供参考======
供参考答案1:
求不定积分∫{(arctanx)/[x²(x²+1)]}dx
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