大一高数定积分

Question

如题所述

Answer 1

本回答被提问者采纳

Answer 2

设 x + 1 = t^2，则 dx = 2t * dt。积分限变换为 t = 0 → 3
原积分变换为：
=∫2t *dt/(t + t^3)
=2*∫dt/(t^2 + 1)
=2*arctan(t)|t = 1 → 3
=2arctan3 - 2arctan1
=2arctan3 - 2*(π/4)
=2arctan3 - π/2

Answer 3

原式=∫dx-∫sin²xsinxdx
=x+∫(1-cos²x)d(cosx)
=[x+cosx-1/3cos³x](0,π)
=[π+cosπ-1/3cos³π]-[0+cos0-1/3cos0]
=π-1+1/3-1+1/3
=π-4/3

Answer 4

令t=sqrt(x+1)，f(t)=2/(1+t^2),这个积分可以再高数积分表中查到。