令1+√x=t,
则x=(t-1)²,
所以
∫ arctan(1+√x)dx
=∫ arctant d[(t-1)²] 使用分部积分法
=(t-1)² *arctant - ∫ (t-1)² d(arctant)
=(t-1)² *arctant - ∫ (t-1)²/(1+t²) dt
=(t-1)² *arctant - ∫ (t²-2t+1)/(1+t²) dt
=(t-1)² *arctant - ∫ 1- 2t/(1+t²) dt
=(t-1)² *arctant - t +ln|1+t²| +C 代入x=(t-1)²
=x *arctan(1+√x) - (1+√x) + ln|x+2√x+2| +C ,C为常数