Question

求不定积分arctan(x^1/2)/((x^1/2)×(1+x))dx

Answer 1

设 u=arctan [x^(1/2)],

则 du = (1/2)x^(-1/2) / (1+x)

∫arctan [x^(1/2)] / [ (1+x) x^(1/2) ]dx = ∫ 2udu = u^2 + C

= { arctan [x^(1/2)] }^2 + C