Question

不定积分∫[(x+1)^1/2-1]/(x+1)^1/2+1]dx ∫1/[(x^2+a^2)^3/2]dx ∫x^1/2/[1-x^1/3]dx 求解题过程

用换元积分法求解

Answer 1

∫[√(x+1)-1]dx/[√(x+1)+1] √(x+1)=t换元
=∫dx-∫2dx/√[(x+1)+1]
=x-2∫d√(x+1)^2/[√(x+1)+1]
=x-4∫√(x+1)d√(x+1)/[√(x+1)+1]
=x-4∫d√(x+1)+4∫d√(x+1)/[√(x+1)+1]
=x-4√(x+1)+4ln|√(x+1)+1|+C

∫dx/√(x^2+a^2)^3
x=atanu sinu=x/√(x^2+a^2)
=∫asecu^2du/a^3secu^3
=(1/a^2)sinu+C
=(1/a^2)(x/√(x^2+a^2))+C

∫x^(1/2)dx/(1-x^(1/3)) x^(1/6)=t 换元
=(2/3)∫dx^(3/2)/(1-x^(1/3))
=(2/3)∫dx^(9/6)/(1-x^(1/3))
=(2/3)∫9x^(8/6)dx^(1/6)/(1-x^(1/3))
=6∫x^(4/3)dx^(1/6)/(1-x^(1/3))
=6∫(x^(4/3)-1)dx^(1/6)/(1-x^(1/3))+6∫dx^(1/6)/(1-x^(1/3))
= 6∫-(1+x^(1/3)(1+x^(2/3))dx^(1/6) +3∫(1+x^(1/6)+1-x^(1/6))dx^(1/6)/[(1-x^(1/6)(1+x^(1/6))
=-6∫1+x+x^(1/3)+x^(2/3)dx^(1/6)+3∫dx^(1/6)/(1-x^(1/6)+3∫dx^(1/6)/(1+x^(1/6))
=-6x^(1/6)-(6/7)x^(7/6)-2x^(1/2)-(6/5)x^(5/6)+3ln|1+x^(1/6)|/|1-x^(1/6)|+C