∫1/x²(1+x²)dx
=∫(1+x²-x²)/x²(1+x²)dx
=∫(1+x²)/x²(1+x²)dx-∫x²/x²(1+x²)dx
=∫1/x²dx-∫1/(1+x²)dx
=-1/x-arctanx+C
∫1/x2(1+x2)dx 求此式子的不定积分~
∫1/x²(1+x²)dx
=∫(1+x²-x²)/x²(1+x²)dx
=∫(1+x²)/x²(1+x²)dx-∫x²/x²(1+x²)dx
=∫1/x²dx-∫1/(1+x²)dx
=-1/x-arctanx+C