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
