How do you integrate #int 1/sqrt(3x^2-12x+29)dx# using trigonometric substitution?

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Answer 1 · 2018-03-16T13:10:45

Use the substitution #3x-6=sqrt51tantheta#.

Let

#I=int1/sqrt(3x^2-12x+29)dx#

Complete the square in the square root:

#I=sqrt3int1/sqrt((3x-6)^2+51)dx#

Apply the substitution #3x-6=sqrt51tantheta#:

#I=sqrt17intsecthetad theta#

Integrate directly:

#I=sqrt17ln|sectheta+tantheta|+C#

Rescale #C# and reverse the substitution:

#I=sqrt17ln|3x-6+sqrt(3x^2-12x+29)|+C#