Question

Integrate dx/sqrt(5-2x+x²)

Answer 1

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

Answer 2

Yes, thats what I want lol.

Answer 3

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?

Answer 4

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

Answer 5

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

Answer 6

Do you know integration by parts?

Answer 7

Yes, I do.

Answer 8

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).

Answer 9

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?

Answer 10

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...\[u=\frac{x-1}{2};du=\frac{1}{2}dx;2du=dx\]\[\int\limits_{}^{}\frac{2du}{2\sqrt{u ^{2}+1}}=\int\limits_{}^{}\frac{1}{\sqrt{u ^{2}+1}}du=\sinh ^{-1}u+C\]\[=\sinh ^{-1}(\frac{x-1}{2})+C\]