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To evaluate this integral, first complete the square (under the square-root sign) in the denominator of the integrand, as suggested, giving:
`I = int 1/((x-1)sqrt((x-1)^2 - 1)) dx`
Notice now that the integrand has the element `1/sqrt(y^2 - a^2)` in it, which suggests that a secant substitution would be sensible.
So use the substitution `x-1 = sec theta` so that `dx = sec theta tan theta d theta`
Now rewrite the integral as
`I = int (sec theta tan theta)/(sec theta sqrt(sec^2 theta - 1)) d theta`
Knowing the trigonometric relation `tan theta = sqrt(sec^2 theta - 1)` , we then have that
`I = int (sec theta tan theta)/ (sec theta tan theta) d theta = int 1. d theta` `= theta` + c
Putting `x - 1 = sec theta` back in, establish `theta` in terms of `x`
`theta = sec^(-1)(x-1) = arcsec (x-1)` where arcsec is the inverse secant function.
Therefore `I = arcsec(x-1) + c` where `c` is a constant of integration.
NB see Example 6 in the first reference link below
You may want to try the online wolfram calculator: http://integrals.wolfram.com/index.jsp?expr=1%2F%28%28x-1%29Sqrt%5Bx%5E2-2x%5D%29&random=false
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