(1)
let
x=2tanu
dx=2(secu)^2 du
∫ dx/[x^2.√(4+x^2)]
=∫ 2(secu)^2 du/[(2tanu)^2.(2secu)]
=(1/4)∫ secu/(tanu)^2 du
=(1/4)∫ cosu/(sinu)^2 du
=(1/4)∫ dsinu/(sinu)^2
=-(1/4)(1/sinu) + C
=-(1/4)[√(4+x^2)/x] + C
(2)
let
x=sinu
dx=cosu du
∫ x/[(x^2+1).√(1-x^2)] dx
=∫ sinu/[(sinu)^2+1] du
=∫ sinu/[2-(cosu)^2] du
=-∫ dcosu/[2-(cosu)^2]
=-(1/2)∫ dcosu/{1-[(cosu)/√2]^2}
=-(√2/2)∫ d(cosu/√2)/{1-[(cosu)/√2]^2}
=-(√2/2)arctan[cosu/√2] + C
=-(√2/2)arctan[√(1-x^2)/√2] + C
追问第二题答案不对