求∫ (2x^2-1)/x(x-4)dx 的不定积分 ∫1/√（4x^2+9）dx的不定积分

Question

如题所述

Answer 1

∫(2x^2-1)dx/[x(x-4)]
=∫[(2x^2-8)+7]dx/[x(x-4)]
=2x+(7/4)∫[x-(x-4)]dx/[x(x-4)]
=2x+(7/4)ln|(x-4)/x|+C

∫dx/√(4x^2+9)
=(1/2)∫d(2x/3)/√[(2x/3)^2+1]
2x/3=tanu √(2x/3)^2+1=secu
sinu=tanu/secu=(2x/3)/√[(2x/3)^2+1]=2x/√(4x^2+9)
=(1/2)∫secu^2du/secu
=(1/2)∫cosudu
=(1/2)sinu+C
=x/√(4x^2+9) +C