For the integrand #1/(sqrt(3x-12sqrt(x)+29))#, substitute
#u=sqrt(x)# and #du=1/(2sqrt(x))dx#
#=2int(u/(sqrt(3u^2-12u+29)))du#
Completing the square, we get
#=2int(u/(sqrt((sqrt(3)u-2sqrt(3))^2+17)))du#
Let #s=sqrt(3)u-2sqrt(3)# and #ds=sqrt(3)du#
#=2/sqrt(3) int ((s+2sqrt(3))/(sqrt(3)sqrt(s^2+17))ds)#
Factor out the constant #1/sqrt(3)#
#=2/3 int(s+2sqrt(3))/(sqrt(s^2+17)) ds#
Next, substitute #s=sqrt(17) tan(p)# and #ds=sqrt(17)sec^2(p)dp#
Then, #sqrt(s^2+17)=sqrt(17tan^2(p)+17)=sqrt(17)sec(p)# and #p=tan^-1(s/sqrt(17))#. This gives us
#=(2sqrt(17))/3 int (sqrt(17)tan(p)+2sqrt(3)sec(p))/sqrt(17) dp#
Factoring out #1/sqrt(17)# gives
#=2/3 int (sqrt(17)tan(p) + 2sqrt(3))sec(p) dp#
#=2/3 int (sqrt(17)tan(p)sec(p) + 2sqrt(3)sec(p)) dp#
#=(2sqrt(17))/3 int (sqrt(17)tan(p)sec(p))dp + 4/sqrt(3) int sec(p)) dp#
Substitute #w=sec(p)# and #dw=tan(p)sec(p)dp#
#=(2sqrt(17))/3 int 1dw + 4/sqrt(3) intsec(p)dp#
#=(2sqrt(17)w)/3+4/sqrt(3) int sec(p)dp#
#=(4 ln(tan(p)+sec(p)))/sqrt(3) + (2sqrt(17)3)/3+C#
Backwards substituting all the variables eventually gives
#=2/3(sqrt(3x-12sqrt(x)+29)+2sqrt(3) sinh^-1(sqrt(3/17)(sqrt(x)-2)))+C#
