Recent content by laker88116

She should call the kid Ariel... get it?

Alright, that makes sense, thanks.

I'm not understanding that. What does it have to do with my equation I listed.

Problem is, I don't know what Gamma is other than a greek letter. I can use the formula, that's not the problem. I just was curious if there was a way to prove it. I was messing with my calculator and I noticed that half numbers have factorials and other decimals don't. So, I looked this up...

Any ideas on how to prove this? (n+\tfrac{1}{2})! = \sqrt{\pi} \prod_{k=0}^{n}\frac{2k+1}{2}

Ok, I think I got it. Thanks.

Ok, I get A=-1, b=1/2, c=1/2, is that right?

I know that, but then what do I do, I have to get a common denominator, which is u(u+1)(u-1) , but then what?

\int \frac {1}{x\sqrt{4x+1}}dx Here's what I have done so far on this problem I let u= \sqrt{4x+1} , so then u^2=4x+1 , du= \frac {2dx}{u} and x= \frac {u^2-1}{4} Substituting, I get \int \frac {1}{(\frac{u^2-1}{4})u}du Then moving stuff around, I get 4 \int \frac...

Alright, I think I got this better now. Thanks.

I understand that and it makes sense. I just don't get how you determine that's what you do.

\int \frac {x+4}{x^2+2x+5} I have no idea where to start on this. I can't see any substitutions that would work. I tried completing the square. I also tried to split up the fraction. It isn't getting any simpler. Any help is appreciated.

Wow, how did I miss that, thanks so much.

\int ln(2x+1)dx So far I know that I need to use integration by parts, I let u= ln(2x+1) and so du= \frac {dx}{2x+1} . Also, I said dv= dx and v=x . So then plugging this into the equation for integration I get: xln(2x+1) - \int \frac {2x}{2x+1}dx Then I determine that I...

Ok, well I understand now I was able to interpret it after VietDao29 did most of the work, but anyways, thanks for the help I see it now.