How do you integrate #int (4x)/sqrt(x^2-14x+53)dx# using trigonometric substitution?

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Answer 1 · 2018-01-20T20:29:09

The answer is #=4sqrt(x^2-4x+53)+28ln((x-7)/2+sqrt(1+(x-7)^2/4))+C#

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#