Question

不定积分arctan(1+x^1/2)dx

Answer 1

∫arctan(1+√x)dx

令√x=t

x=t^2

dx=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

=xarctan(1+√x)-√x+arctan√x +C