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...

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- Oct 27th 2006, 10:12 PMscorpion007another integral
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... - Oct 27th 2006, 10:54 PMCaptainBlack
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 - Oct 27th 2006, 11:13 PMscorpion007
is this correct?

$\displaystyle 2\arcsin(\frac{x}{2}) - 6\sqrt{4-x^2}$ - Oct 28th 2006, 12:39 AMCaptainBlack
- Oct 28th 2006, 07:22 AMSoroban
Hello, scorpion007!

This is the Captain's solution . . . in baby-talk.

Quote:

$\displaystyle \int \frac{2+6x}{\sqrt{4-x^2}}\,dx$

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 $\displaystyle u = 4-x^2$

- Oct 28th 2006, 07:44 PMscorpion007
thanks, i understand perfectly.