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Eleven17Best ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}\frac{ 1 }{ \sqrt{5  2x + x^{2}} }\]
 one year ago

henriquerbBest ResponseYou've already chosen the best response.0
Yes, thats what I want lol.
 one year ago

Eleven17Best ResponseYou've already chosen the best response.0
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?
 one year ago

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

Eleven17Best ResponseYou've already chosen the best response.0
No problem! I love doing Calculus...such a nerd.
 one year ago

Eleven17Best ResponseYou've already chosen the best response.0
Do you know integration by parts?
 one year ago

EulerGroupieBest ResponseYou've already chosen the best response.1
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).
 one year ago

henriquerbBest ResponseYou've already chosen the best response.0
EulerGroupie, 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?
 one year ago

EulerGroupieBest ResponseYou've already chosen the best response.1
Yes, 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...\[u=\frac{x1}{2};du=\frac{1}{2}dx;2du=dx\]\[\int\limits_{}^{}\frac{2du}{2\sqrt{u ^{2}+1}}=\int\limits_{}^{}\frac{1}{\sqrt{u ^{2}+1}}du=\sinh ^{1}u+C\]\[=\sinh ^{1}(\frac{x1}{2})+C\]
 one year ago

KrishnadasBest ResponseYou've already chosen the best response.0
dw:1355917276626:dw
 one year ago

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

KrishnadasBest ResponseYou've already chosen the best response.0
@EulerGroupie is \[\log_{}x+\sqrt{x ^{2}+a ^{2}} ?\]
 one year ago

EulerGroupieBest ResponseYou've already chosen the best response.1
\[\ln(x+\sqrt{x ^{2}+1})=\sinh ^{1}(x)\]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.
 one year ago

KrishnadasBest 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..
 one year ago
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