Can someone check this integration?

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Can someone check this integration?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Other answers:

So if you say |dw:1434570936513:dw|
That is after IBP
Then can't you do a sub?
So it looks like you set \(u=\ln x\) and \(dv=\dfrac{x}{\sqrt{x^2-1}}\,dx\). A trig substitution might work. Try \(x=\sec u\).
That give|dw:1434571156177:dw|
Let's see... \[\int\frac{\sqrt{x^2-1}}{x}\,dx=\int\frac{\sqrt{\sec^2u-1}}{\sec u}\sec u\tan u\,du=\int \tan^2u\,du\] Now a trig identity will work well here: \[\tan^2u=\sec^2u-1\] and you know the antiderivative of \(\sec^2u\).
Yes, so does that not give
|dw:1434571377468:dw|
That's right.
Moving on then, I end up at the end getting something different :(|dw:1434571498405:dw|
\(\ln x(x-1)\) should be \((\ln x-1)\).
Whoops! Ok. But does that match any of those possible answers, or have we made a mistake?
Other than that, your solution is right. I suspect a typo in the answer choices.
Thanks. I wonder if any are equivalent?
Because I have not ever had a problem where they have made a mistake before.
You can differentiate the answer choices to check. The second one can be eliminated right away since it doesn't generate any log terms.
Ok. Thanks so much!
You're welcome!

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