Question

A hard integral. Anyone? :) (Note: Just ping if you already gave up, i have solution. :))

Answer 1

Oh man is that a jerk to look at...but lets see if a simple u-sub would work

\[\large u = \sqrt{1 + \sqrt{x} } + 1\]
so then du would be \(\large \frac{1}{4\sqrt{1 + \sqrt{x}}\sqrt{x}}\)

And if that is true we would have

\[\large 4\int^{1}_{0} \frac{1}{\sqrt{u}}du\]

which of course is \[\large 2\sqrt{u}\] so we would have \[\large 8\sqrt{u}\] which would be

\[\large 8\sqrt{\sqrt{1 + \sqrt{x}} + 1}\]

And that from 0 to 1 would be...

\[\large 8\sqrt{\sqrt{1 + \sqrt{1}} + 1} - 8\sqrt{\sqrt{1 + \sqrt{0}} + 1}\] which if my math is correct anI didnt miss a square root lol...should be \(\large 12.4 - 11.3\) which is \(\large \approx 1.1\)

Look about right?

Answer 2

Yes you got it. :)

Answer 3

WOO! haha thank you calculus!