凑微分法，将1╱(1+x²)dx变成d(arctanx)，原式为1/3arctan³x＋C
y'=arctanx+x/(1+x²)dy=[arctanx+x/(1+x²)]dxlny=xln(1+x²)两边同时对x求导,得y'/y=ln(1+x²)+2x²/(1+x²)y'=(1+x^2)^x[ln(1+x²)+2x²/(1+x²)]所以dy=(1+x^2)^x[ln(1+x²)+2x²/(1+x²)]dx
