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October 6th, 2018, 05:18 PM   #11
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You slipped up when doing the initial substitution.

Also, you forgot to include a constant of integration.
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October 6th, 2018, 07:31 PM   #12
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Apart from skipjack's comments, it looks good. But if you are going to use a substitution $x=\sec u$, why not do it on the integral $\displaystyle \int \sqrt{x^2-1} \,\mathrm dx$? Admittedly it leads to the mildly challenging $$\int \sec^3 u \, \mathrm du$$
I approached that one via \begin{align}\int \sec{(u)} \, \mathrm du &= \int \cos^2{(u)}\sec^3{(u)}\, \mathrm du \\ &= \int (1-\sin^2{(u)})\sec^3{(u)}\, \mathrm du \\ &= \int \sec^3{(u)}\,\mathrm du - \int \underbrace{\sin{(u)}}_{u} \underbrace{\sin{(u)} \cos^{-3}{(u)}}_{\mathrm dv}\, \mathrm du \end{align}
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October 6th, 2018, 08:05 PM   #13
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Substituting $x = \cosh u$ is simpler for that (or for integrating $x^2 /\sqrt{x^2 - 1}$) if you know about the hyperbolic functions.
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