## anonymous 3 years ago Integrate dx/sqrt(5-2x+x²)

1. anonymous

$\int\limits_{}^{}\frac{ 1 }{ \sqrt{5 - 2x + x^{2}} }$

2. anonymous

Yes, thats what I want lol.

3. anonymous

Haha, Okay! I was going to be rude and just post that picture of it, but I'll actually solve it out for you, do you want any of the work shown?

4. anonymous

Hahaha, yes, I can't solve that one. You knw when your brain is blocked? If you could do that one I would apreciate.

5. anonymous

No problem! I love doing Calculus...such a nerd.

6. anonymous

Do you know integration by parts?

7. anonymous

Yes, I do.

8. anonymous

I came up with two methods. Both started with completing the square under the radical. After that you can either do a trig. sub. (the long way), or do a u sub. resulting in an inverse hyperbolic trig function (the short way).

9. anonymous

EulerGroupie, x²-2x+5=(x-1)²+4, is that what you mean by completing squares? (I believe so). I can complete squares, but I'm always confused with that 4 in the sub's. Could you help me?

10. anonymous

Yes, you completed the square properly.$\int\limits_{}^{}\frac{1}{\sqrt{(x-1)^{2}+4}}dx$The first step is to recognize that we have an integral that looks like the generalized formula...$\sinh ^{-1}u+C=\int\limits_{}^{}\frac{1}{\sqrt{u ^{2}+1}}du$Then we use algebra to get it closer to this form by factoring a 4 from the radicand and eventually taking it out of the radical as its root...$\int\limits_{}^{}\frac{1}{\sqrt{4[\frac{(x-1)^{2}}{4}+1]}}dx=\int\limits_{}^{}\frac{1}{2\sqrt{(\frac{x-1}{2})^{2}+1}}dx$Notice how the four left inside is incorporated into the squared term. Now we can use a u-sub where...$u=\frac{x-1}{2};du=\frac{1}{2}dx;2du=dx$$\int\limits_{}^{}\frac{2du}{2\sqrt{u ^{2}+1}}=\int\limits_{}^{}\frac{1}{\sqrt{u ^{2}+1}}du=\sinh ^{-1}u+C$$=\sinh ^{-1}(\frac{x-1}{2})+C$

11. anonymous

|dw:1355917276626:dw|

12. anonymous

now you can do..

13. anonymous

@EulerGroupie isnt $\int\limits_{}^{} dx/\sqrt{x ^{2}+a ^{2}}$

14. anonymous

@EulerGroupie is $\log_{}|x+\sqrt{x ^{2}+a ^{2}} ?$

15. anonymous

$\ln(x+\sqrt{x ^{2}+1})=\sinh ^{-1}(x)$I came across the log solution as well, but when I tried to double check my answer with Wolfram, it wasn't there. It offered up the inverse hyperbolic sine solution so I reviewed that section in my Calc text and sure enough, they are equivalent. I thought the the sinh seemed more elegant (and verifiable through Wolfram), so I went with it.

16. anonymous

@EulerGroupie when the lesson on integration was taken on my class,we studied only that equation...thats why...there must be many answers to one question in integration...we both are correct i think..

17. anonymous

absolutely