求不定积分∫{(arctan根号下x)/[(根号下x)(1+x)]}dx

Question

求不定积分 ∫ { (arctan根号下x) / [ (根号下x) (1+x) ] }dx

Answer 1

令根号下x=t,x=t^2∫ { (arctan根号下x) / [ (根号下x) (1+x) ] }dx=∫ { (arctan t / [ t (1+t^2) ] }d(t^2)=∫ { (arctan t / [ t (1+t^2) ] }2tdt=∫2 { (arctan t / (1+t^2) }dt =∫ 2arctan t d(arctan t)=(arctan t)^2+C=(arctan 根号下x)^2+C