计算不定积分∫(1+√x)^2dx/√x

题目

计算不定积分∫(1+√x)^2dx/√x

答案

∫ (1 + √x)² / √x dx
= ∫ (1 + 2√x + x) / √x dx
= ∫ (1 / √x + 2 + √x) dx
= [2√x + 2x + (2/3)x^(3/2)] + C
= (2/3)√x(x + 3√x + 3) + C