Let
(c) Evaluate the integral for f ( x , y ) = 1 ( 1 + y ) 2 {\displaystyle f(x,y)={\frac {1}{(1+y)^{2}}}}
Hint: You may use 1 9 − x 2 = 1 6 ( 1 x + 3 − 1 x − 3 ) {\displaystyle {\frac {1}{9-x^{2}}}={\frac {1}{6}}\left({\frac {1}{x+3}}-{\frac {1}{x-3}}\right)}