المسالة بالمرفق

 

 

(1)= integral sqrt (x/9-x)

let u -sqrt(x /9-x) then dx =18/ ( u^2+ 1)^2 *udu
so (1)= integral 18u^2/(u^2+1)^2 du

let u= tan z then du sec^2 z dz

then(1)=18 integral sin^2 z dz = 9 integral (1 - cos 2 z ) dz
=9( z -1/2 sin 2 z )

=9(tan^-1 u - u/ u^2+1 )

=9(tan^-1 sqrt(x/9-x) -sqrt (x/9-x) / 9//9-x )



the required integral =9(tan^-1 sqrt 1/8 - (sqrt 1/8 ) /(9/8 )


- u/u^2+1