∫1/(1+x^²)²dx 求解

Answer 1

如图所示

Answer 2

∫1/(1+x²)²dx =1/2∫1/[x(1+x²)²]d(1+x²)
=-1/2∫1/xd]1/(1+x²)]
=-1/[2x(1+x²)]+1/2∫1/(1+x²)d1/x
=-1/[2x(1+x²)]-1/2∫1/[x²(1+x²)]dx
=-1/[2x(1+x²)]-1/2∫1/x²-1/(1+x²)dx
=-1/[2x(1+x²)]+1/(2x)-arctanx/2+C

Answer 3

设x＝tana
sina＝x/√(1+x^2)，cosa＝1/√(1+x^2)
原式＝∫cos^2ada＝a/2+1/2sinacosa＝1/2arctanx+x/[2(1+x^2)]+C