计算积分 ∫[sinx+(x^2)]dx ∫xe^[(-x^2)]dx ∫{1/[根号（x+1)]}dx

题目

计算积分 ∫[sinx+(x^2)]dx ∫xe^[(-x^2)]dx ∫{1/[根号（x+1)]}dx
∫[sinx+(x^2)]dx
∫xe^[(-x^2)]dx
∫{1/[根号（x+1)]}dx

答案

∫[sinx+(x^2)]dx=∫ sinx dx + ∫ x²dx= -cosx + (1/3)x³ + C∫xe^[(-x²)]dx=(-1/2)∫ [e^(-x²)] d(-x²)=(-1/2) [e^(-x²)] + C∫{1/[√（x+1)]}dx=∫{1/[√（x+1)]}d(x+1)= 2 √(x+1...