#\ \ \ \ \ \ int_0^1 9/(3+x^2)^2\ dx#
Substitute #x=sqrt(3)tan(theta)# and #dx=sqrt(3)sec^2(theta)d\theta#:
#=int_arctan(0)^arctan(1/sqrt(3))9/(3+(sqrt(3)tan(theta))^2)^2sqrt(3)sec^2(theta)\ d\theta#
#=sqrt(3)int_0^(pi/6) (9sec^2(theta))/(3+3tan^2(theta))^2\ d\theta#
#=sqrt(3)int_0^(pi/6) sec^2(theta)/(1+tan^2(theta))^2\ d\theta#
Using the fact that #tan^2(theta)+1=sec^2(theta)#:
#=sqrt(3)int_0^(pi/6) sec^2(theta)/sec^4(theta)\ d\theta#
#=sqrt(3)int_0^(pi/6) 1/sec^2(theta)\ d\theta#
Since #sec(theta)=1/cos(theta)#, we have
#=sqrt(3)int_0^(pi/6) cos^2(theta)\ d\theta#
Using the fact that #cos^2(theta)=(1+cos(2theta))/2#:
#=sqrt(3)/2int_0^(pi/6) 1+cos(2theta)\ d\theta#
#=sqrt(3)/2[theta+sin(2theta)/2]_0^(pi/6)#
#=sqrt(3)/2(pi/6+1/4)#
#=((2pi+3)sqrt(3))/24#
