another integral

**scorpion007** · October 27th 2006, 10:12 PM

Find an antiderivative of $\frac{2+6x}{\sqrt{4-x^2}}$

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...

**CaptainBlack** · October 27th 2006, 10:54 PM

Originally Posted by scorpion007

Find an antiderivative of $\frac{2+6x}{\sqrt{4-x^2}}$

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:

$\frac{d}{dx} \sqrt{4-x^2} = \frac{-x}{\sqrt{4-x^2}}$ .

So if we rewrite the problem as find the antiderivative of:

$\frac{2+6x}{\sqrt{4-x^2}} = \frac{2}{\sqrt{4-x^2}} + \frac{6x}{\sqrt{4-x^2}}$

as the first term on the right is a standard integral, and the second is $-6$ times what we had earlier you should be able to complete this from here.

RonL

**scorpion007** · October 27th 2006, 11:13 PM

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

**CaptainBlack** · October 28th 2006, 12:39 AM

Originally Posted by scorpion007

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

Looks OK to me.

Checking by differentiation is shown in the attachment.

RonL

**Soroban** · October 28th 2006, 07:22 AM

Hello, scorpion007!

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

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

We have: . $\int\frac{2+6x}{\sqrt{4-x^2}}\,dx \;=\;\int\left(\frac{2}{\sqrt{4-x^2}} + \frac{6x}{\sqrt{4-x^2}}\right)dx <br />$

. . . . . $=\;2\!\int\!\!\frac{dx}{\sqrt{4-x^2}} \;+ \;6\!\int\!\! x(4-x^2)^{-\frac{1}{2}}dx$

The first integral is of the $\arcsin$ form.

For the second integral, let $u = 4-x^2$

**scorpion007** · October 28th 2006, 07:44 PM

thanks, i understand perfectly.

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