What is the integral of 1/(4x + 8 (square root of x)?

Mr. Daftary's solution is the quickest & best way to solve this (BA), but there should be "8" in the given :)

So I'll just copy his solution, appending "8" in the denominator

∫dx / (4x + 8√x)

Let √x = t => x = t^2 => dx = 2t dt

=> Integral

= 2∫t dt / (4t^2 + 8t)

= (1/2) ∫dt / (t + 2)

= (1/2) ln (t + 2) + c

= (1/2) ln (√x + 2) + c.

Now if someone could explain to me why I got a different answer...

http://answers.yahoo.com/question/index?qid=200910...

∫dx / (4x + √x)

Let √x = t => x = t^2 => dx = 2t dt

=> Integral

= 2∫t dt / (4t^2 + t)

= 2∫dt / (4t + 1)

= 2 * (1/4) ln (4t + 1) + c

= (1/2) ln (4√x + 1) + c.

Wolfram|Alpha is a great resource for difficult integrals :)

http://www.wolframalpha.com/input/?i=Integrate%5B1...

Click the show steps button to see exactly how to do it.

Edit: .5*ln(sqrt(x)+2)+C is the final answer