A community for students.
Here's the question you clicked on:
 0 viewing
henriquerb
 4 years ago
Integrate dx/sqrt(52x+x²)
henriquerb
 4 years ago
Integrate dx/sqrt(52x+x²)

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}\frac{ 1 }{ \sqrt{5  2x + x^{2}} }\]

henriquerb
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, thats what I want lol.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Haha, 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?

henriquerb
 4 years ago
Best ResponseYou've already chosen the best response.0Hahaha, yes, I can't solve that one. You knw when your brain is blocked? If you could do that one I would apreciate.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No problem! I love doing Calculus...such a nerd.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you know integration by parts?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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).

henriquerb
 4 years ago
Best ResponseYou've already chosen the best response.0EulerGroupie, x²2x+5=(x1)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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, you completed the square properly.\[\int\limits_{}^{}\frac{1}{\sqrt{(x1)^{2}+4}}dx\]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...\[\int\limits_{}^{}\frac{1}{\sqrt{4[\frac{(x1)^{2}}{4}+1]}}dx=\int\limits_{}^{}\frac{1}{2\sqrt{(\frac{x1}{2})^{2}+1}}dx\]Notice how the four left inside is incorporated into the squared term. Now we can use a usub where...\[u=\frac{x1}{2};du=\frac{1}{2}dx;2du=dx\]\[\int\limits_{}^{}\frac{2du}{2\sqrt{u ^{2}+1}}=\int\limits_{}^{}\frac{1}{\sqrt{u ^{2}+1}}du=\sinh ^{1}u+C\]\[=\sinh ^{1}(\frac{x1}{2})+C\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1355917276626:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@EulerGroupie isnt \[\int\limits_{}^{} dx/\sqrt{x ^{2}+a ^{2}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@EulerGroupie is \[\log_{}x+\sqrt{x ^{2}+a ^{2}} ?\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\ln(x+\sqrt{x ^{2}+1})=\sinh ^{1}(x)\]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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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..
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.