Let's rewrite the function
(4x)/sqrt(x^2-14x+53)
=2((2x-14)/sqrt(x^2-14x+53))+2(14/sqrt(x^2-14x+53))
Therefore, there are 2 integrals
int(4xdx)/sqrt(x^2-14x+53)=2int(((2x-14)dx)/sqrt(x^2-14x+53))+2int((14dx)/sqrt(x^2-14x+53))
Perform the integrals separately,
2int(((2x-14)dx)/sqrt(x^2-14x+53))=4sqrt(x^2-4x+53)
2int((14dx)/sqrt(x^2-14x+53))=28int(dx)/sqrt(x^2-14x+53)
Complete the squares in the denominator
x^2-14x+53=x^2-14x+49+53-49=(x-7)^2+4=4(((x-7)/2)^2+1)
Let u=(x-7)/2, =>, du=dx/2
28int(dx)/sqrt(x^2-14x+53)=28int(dx)/(sqrt(4(((x-7)/2)^2+1)))
=14int(2du)/sqrt(u^2+1)
=28int(du)/(sqrt(u^2+1))
Let u=tantheta, =>, du=sec^2thetad theta
1+tan^2theta=sec^2theta
28int(du)/(sqrt(u^2+1))=28int(sec^2theta d theta)/(sectheta)
=28intsectheta d theta
=28ln(tan theta + sectheta)
=28ln(u+sqrt(1+u^2))
=28ln((x-7)/2+sqrt(1+(x-7)^2/4))
Finally,
int(4xdx)/sqrt(x^2-14x+53)=4sqrt(x^2-4x+53)+28ln((x-7)/2+sqrt(1+(x-7)^2/4))+C
