## AmTran_Bus one year ago Can someone check this integration?

1. AmTran_Bus

2. anonymous

What have you tried? By parts?

3. AmTran_Bus

4. AmTran_Bus

So if you say |dw:1434570936513:dw|

5. AmTran_Bus

That is after IBP

6. AmTran_Bus

Then can't you do a sub?

7. anonymous

So it looks like you set $$u=\ln x$$ and $$dv=\dfrac{x}{\sqrt{x^2-1}}\,dx$$. A trig substitution might work. Try $$x=\sec u$$.

8. AmTran_Bus

That give|dw:1434571156177:dw|

9. anonymous

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 $$\sec^2u$$.

10. AmTran_Bus

Yes, so does that not give

11. AmTran_Bus

|dw:1434571377468:dw|

12. anonymous

That's right.

13. AmTran_Bus

Moving on then, I end up at the end getting something different :(|dw:1434571498405:dw|

14. anonymous

$$\ln x(x-1)$$ should be $$(\ln x-1)$$.

15. AmTran_Bus

Whoops! Ok. But does that match any of those possible answers, or have we made a mistake?

16. anonymous

Other than that, your solution is right. I suspect a typo in the answer choices.

17. AmTran_Bus

Thanks. I wonder if any are equivalent?

18. AmTran_Bus

Because I have not ever had a problem where they have made a mistake before.

19. anonymous

You can differentiate the answer choices to check. The second one can be eliminated right away since it doesn't generate any log terms.

20. AmTran_Bus

Ok. Thanks so much!

21. anonymous

You're welcome!