∫(0->; 1) dx/(x^2-x-2)
=∫(0->; 1) dx/[(x-2)(x+ 1)]
=( 1/3)∫(0->; 1)[ 1/(x-2)- 1/(x+ 1)]dx
=( 1/3)[ln |(x-2)/(x+ 1)|]|(0-& gt; 1)
=( 1/3)[ln( 1/2)-LN2]
=-(2/3)ln2
(2)
let
u= (arctan√x)^2
du = 2 arctan√x .[ 1/( 1+x)]
= {arctan√x /[√x.( 1+x) ] } dx
∫(0->; +∞) arctan√x /[ √x. ( 1+x) ] dx
=[(arctan√x)^2]|(0->; +∞)
=( 1/4)π^2