## henriquerb Group Title Integrate dx/sqrt(5-2x+x²) one year ago one year ago

1. Eleven17

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

2. henriquerb

Yes, thats what I want lol.

3. Eleven17

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. henriquerb

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. Eleven17

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

6. Eleven17

Do you know integration by parts?

7. henriquerb

Yes, I do.

8. EulerGroupie

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. henriquerb

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. EulerGroupie

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$

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now you can do..

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

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

15. EulerGroupie

$\ln(x+\sqrt{x ^{2}+1})=\sinh ^{-1}(x)$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.