解答过程如下图:
令√x=t,x=t²,则dx=2tdt原式=∫arctant*2tdt/t(1+t²)=2∫arctantdt/(1+t²)=2∫arctantd(arctant)=2[arctan²t-∫arctantd(arctant)]即原式=2∫arctantd(arctant)=arctan²t∴原式=1/2*arctan²t+C=1/2*arctan²√x+C
