不定积分arctan(1+x^1/2)dx

题目

不定积分arctan(1+x^1/2)dx

答案

∫arctan(1+√x)dx令√x=tx=t^2dx=dt^2原式化为∫arctan(1+t)*dt^2=t^2arctan(1+t)-∫t^2*1/(1+t^2) dt=t^2arctan(1+t)-∫(t^2+1-1)/(t^2+1)dt=t^2arctan(1+t)-∫dt+∫dt/(t^2+1)=t^2arctan(1+t)-t+arctant+C=xarcta...