How do you integrate #int (4x)/sqrt(x^2-14x+65)dx# using trigonometric substitution?

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Answer 1 · 2016-03-08T20:44:29

#4(sqrt(x^2-14x+65) +7sinh^-1((x-7)/4))+ C#

The two main tricks are:

Using 1. we choose our substitution like this:

To get the following:

#(x-7)^2 +4^2 -> 4^2sinh^2u + 4^2= 16cosh^2u#

Which will simplify #sqrt(x^2-14x+65) -> 4cosh u#

We can proceed:
#int (4x)/sqrt(x^2-14x+65)dx = int (4*(4sinh u +7))/(4cosh u)(4 cosh u) du#
#= 4int (4sinh u +7)du#
#= 4*(4cosh u +7u)+ c#

We can then substitute it back in x:

#4(sqrt(x^2-14x+65) +7sinh^-1((x-7)/4))+ C#