∫(x-1)/(x^2+ 2x+ 3) dx
=(1/2) ∫(2x+2)/(x^2+ 2x+ 3) dx - 2∫dx/(x^2+ 2x+ 3)
=(1/2)ln|x^2+ 2x+ 3| - 2∫dx/(x^2+ 2x+ 3)
=(1/2)ln|x^2+ 2x+ 3| - √2 .arctan[(x+1)/√2] + C
---------
consider
x^2+2x+3 = (x+1)^2 +2
let
x+1 =√2tanu
dx =√2(secu)^2 du
∫dx/(x^2+ 2x+ 3)
=∫√2(secu)^2 du/(2(secu)^2)
=(√2/2) ∫ du
=(√2/2) u + C
=(√2/2) arctan[(x+1)/√2] + C
(1-x)^1/3=t,1-x=t^3,dx=-3t^2dt
∫x^2(1-x)^1/3dx
=∫(1-t^3)(-3t^3)dt
=3∫(t^6-t^3)dt
=3(t^7/7-t^4/4)+C
=3t^4(t^3/7-1/4)+C
把(1-x)^1/3=t代入即可
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