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#
