Question

求解答 在线等





Answer 1

(1)
let
x= asinu
dx=acosu du

∫x^2/ √(a^2-x^2) dx
=∫{ (asinu)^2/ (acosu) } [acosu du]
=a^2.∫ (sinu)^2 du
=(1/2)a^2.∫ (1-cos2u) du
=(1/2)a^2.(u -(1/2)sin2u) +C
=(1/2)a^2.(arcsin(x/a)- x.√(a^2-x^2) /a^2 ) +C

(2)
let
u =√(x+1)
du = dx/[2√(x+1)]
dx=2u du

∫[√(x+1) -1]/[√(x+1) +1] dx
=∫[(u-1)/(u+1) ] (2udu)
=2∫ [ u - 2u/(u+1) ] du
=2∫ [ u - 2 + 2/(u+1) ] du
=2[ (1/2)u^2 - 2u +2ln|u+1| ] +C
= u^2 - 4u +4ln|u+1| +C
=(x+1) -4√(x+1) +4ln|√(x+1)+1| +C

(3)
∫x^2/√(2-x) dx

=-2∫x^2 d√(2-x)
=-2x^2.√(2-x) +4∫x.√(2-x) dx
=-2x^2.√(2-x) -(8/3)∫x d(2-x)^(3/2)
=-2x^2.√(2-x) -(8/3)x.(2-x)^(3/2) +(8/3)∫(2-x)^(3/2) dx
=-2x^2.√(2-x) -(8/3)x.(2-x)^(3/2) -(16/15)(2-x)^(5/2) dx +C