∫(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