Question

求不定积分第一题方法一划线部分怎么得来的

Answer 1

如图所示

Answer 2

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