1. ## another 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...

2. Originally Posted by scorpion007
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

3. is this correct?
$\displaystyle 2\arcsin(\frac{x}{2}) - 6\sqrt{4-x^2}$

4. Originally Posted by scorpion007
is this correct?
$\displaystyle 2\arcsin(\frac{x}{2}) - 6\sqrt{4-x^2}$
Looks OK to me.

Checking by differentiation is shown in the attachment.

RonL

5. 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 $\displaystyle u = 4-x^2$

6. thanks, i understand perfectly.