不定积分∫arctan根号x/根号x*1/(1+x)dx

∫(arctan√x)/[√x(1+x)] dx
=∫(arctan√x)/(1+x) d(2√x)
=2∫(arctan√x)/[1+(√x)²] d(√x)
=2∫arctan√x d(arctan√x),where ∫dx/(1+x²)=arctanx+C
=2*(1/2)(arctan√x)²+C
=(arctan√x)²+C本回答被提问者采纳

令 t = 根号x 带入可简便计算
结果 ＝ (arctan(根号x))^2 + C

设x=tan²t