Originally Posted by

**Vamz** $\displaystyle

\displaystyle \int \frac{1}{\sqrt{12 + 4 x - x^2}}dx

$

Can someone please point out where I went wrong?

First, I complete the square:

$\displaystyle

\displaystyle \int \frac{1}{\sqrt{(x-2)^2-16}}dx

$

Factor out a 16:

$\displaystyle

\displaystyle \int \frac{1}{\sqrt{16(\frac{(x-2)^2}{16}-1)}}dx

$

$\displaystyle

\displaystyle \frac{1}{4}\int \frac{1}{\sqrt{\frac{(x-2)^2}{16}-1}}dx

$

Let U = $\displaystyle \frac{x-2}{4}$ & dU=1\4dx :: dx=4du

$\displaystyle

\displaystyle \frac{1}{4}\int \frac{4du}{\sqrt{U^2-1}}dU

$

$\displaystyle

\displaystyle \int \frac{du}{\sqrt{U^2-1}}dU

$

Which is:

$\displaystyle

arccosh(U) = arccosh(\frac{x-2}{4})

$

It seems like my answer is wrong however. Where did I go wrong?

Thanks!