Start off by rewriting as follows
int (2x)/(9-x^2)^(1/2)dx-int 3/(9-x^2)^(1/2)dx∫2x(9−x2)12dx−∫3(9−x2)12dx
For the first integral we use a u substitution
Let u=9-x^2u=9−x2 then (du)/dx=-2xdudx=−2x
dx=(du)/(-2x)dx=du−2x
Make the substitution
-int (2x)/u^(1/2)(du)/(2x)=-int u^(-1/2)du−∫2xu12du2x=−∫u−12du
Integrating we have
-2u^(1/2)−2u12
Back substituting for u
-2(9-x^2)^(1/2)−2(9−x2)12
-2sqrt(9-x^2)−2√9−x2
For the second integral we will use a trigonometric substitution
Let x=3sinthetax=3sinθ the
dx=3costhetad thetadx=3cosθdθ
Make the substitution
int 3/(9-9sin^2theta)^(1/2)3cosd theta∫3(9−9sin2θ)123cosdθ
int (9costheta)/(9[1-sin^2theta])^(1/2)d theta∫9cosθ(9[1−sin2θ])12dθ
3int costheta/(cos^2theta)^(1/2)d theta3∫cosθ(cos2θ)12dθ
3int costheta/costhetad theta3∫cosθcosθdθ
3int d theta3∫dθ
Integrating we get thetaθ
Now theta=arcsin(x/3)θ=arcsin(x3)
So we have 3arcsin(x/3)3arcsin(x3)
Putting the two results together we have
-2sqrt(9-x^2)-3arcsin(x/3)+C−2√9−x2−3arcsin(x3)+C
