求不定積分∫(arctanx)/(x^2(x^2 1))dx網友廻答：
- 求不定積分∫{(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
網友廻答：
- ∫(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(...
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