Find an antiderivative of $\displaystyle \frac{2+6x}{\sqrt{4-x^2}} $
Now I know it has something to do with arcsin, but I'm not exactly sure what i can do to clean up that numerator...
Note that:
$\displaystyle \frac{d}{dx} \sqrt{4-x^2} = \frac{-x}{\sqrt{4-x^2}}$.
So if we rewrite the problem as find the antiderivative of:
$\displaystyle \frac{2+6x}{\sqrt{4-x^2}} = \frac{2}{\sqrt{4-x^2}} + \frac{6x}{\sqrt{4-x^2}}$
as the first term on the right is a standard integral, and the second is $\displaystyle -6$ times what we had earlier you should be able to complete this from here.
RonL
Hello, scorpion007!
This is the Captain's solution . . . in baby-talk.
$\displaystyle \int \frac{2+6x}{\sqrt{4-x^2}}\,dx$
We have: .$\displaystyle \int\frac{2+6x}{\sqrt{4-x^2}}\,dx \;=\;\int\left(\frac{2}{\sqrt{4-x^2}} + \frac{6x}{\sqrt{4-x^2}}\right)dx
$
. . . . . $\displaystyle =\;2\!\int\!\!\frac{dx}{\sqrt{4-x^2}} \;+ \;6\!\int\!\! x(4-x^2)^{-\frac{1}{2}}dx $
The first integral is of the $\displaystyle \arcsin$ form.
For the second integral, let $\displaystyle u = 4-x^2$