- #1

- 1,196

- 0

y=x^(3/2)/3-x^(1/2); [4,16]

And when I was trying to evaluate

integral[4,16] sqrt(1/2+x/4+1/(4x))dx

I had no idea were to begin so I plugged this into wolfram alpha

http://www.wolframalpha.com/input/?i=integral+sqrt(1/2+x/4+1/(4x))dx

and followed it up to this step

1/2*integral[4,16] sqrt(x+1/x+2)dx

It then says it simplified the powers and that

1/2*integral[4,16] sqrt(x+1/x+2)dx = 1/2*integral[4,16] (x+1)/sqrt(x)dx

I don't follow how this was done at all, apparently

sqrt(x+1/x+2) = (x+1)/sqrt(x)

and I don't see how these are equal

I figuered I would post this here because I'm not having problems with the actual calculus but the algebra II, (I think what ever was done to establish that these two expressions are equal to each other is an algebra II topic.)

Thanks for any help you can proivde