∫(0->√3) dx/(9-x^2)1/(9-x^2) =(1/6) [1/(3-x)- 1/(3+x)]∫(0->√3) dx/(9-x^2)=(1/6)∫(0->√3) [1/(3-x)- 1/(3+x)] dx=(1/6)[ln|1/(9-x^2)|]|(0->√3)=(1/6)[ ln(1/6) - 1n(1/9) ]=(1/6)( ln9 - ln6)
