Integrate dx/sqrt(5-2x+x²)

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Integrate dx/sqrt(5-2x+x²)

OCW Scholar - Single Variable Calculus
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[\int\limits_{}^{}\frac{ 1 }{ \sqrt{5 - 2x + x^{2}} }\]
Yes, thats what I want lol.
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?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Hahaha, yes, I can't solve that one. You knw when your brain is blocked? If you could do that one I would apreciate.
No problem! I love doing Calculus...such a nerd.
Do you know integration by parts?
Yes, I do.
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).
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?
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...\[u=\frac{x-1}{2};du=\frac{1}{2}dx;2du=dx\]\[\int\limits_{}^{}\frac{2du}{2\sqrt{u ^{2}+1}}=\int\limits_{}^{}\frac{1}{\sqrt{u ^{2}+1}}du=\sinh ^{-1}u+C\]\[=\sinh ^{-1}(\frac{x-1}{2})+C\]
|dw:1355917276626:dw|
now you can do..
@EulerGroupie isnt \[\int\limits_{}^{} dx/\sqrt{x ^{2}+a ^{2}}\]
@EulerGroupie is \[\log_{}|x+\sqrt{x ^{2}+a ^{2}} ?\]
\[\ln(x+\sqrt{x ^{2}+1})=\sinh ^{-1}(x)\]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.
@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..
absolutely

Not the answer you are looking for?

Search for more explanations.

Ask your own question