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...