integral of 1/(9 + 4x^2)

The book says u should be 2/3x. Can someone enlighten me on why this is? I amdumbfounded :(

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- Feb 22nd 2009, 04:49 AMTYTYconfused with u-subs problem
integral of 1/(9 + 4x^2)

The book says u should be 2/3x. Can someone enlighten me on why this is? I am**dumb**founded :( - Feb 22nd 2009, 04:54 AMmr fantastic
- Feb 22nd 2009, 05:02 AMTYTY
please be patient with me here as I still don't understand this. (2x/3)^2 x 9 would have to equal 4x^2 again here right? but wouldn't it be (18x/3)^2 = 6x^2?

edit: I guess because of the square, you simplify using square root or something. The calculator agrees with your result but somehow it still goes over my head :( - Feb 22nd 2009, 05:15 AMmr fantastic
$\displaystyle 9 \cdot \left( \frac{2x}{3} \right)^2 = 9 \cdot \frac{2x}{3} \cdot \frac{2x}{3} = 9 \cdot \frac{4x^2}{9} = 4x^2$.

If you're studying calculus it's expected you can do this. And do it easily.

I suggest you extensively revise basic algebra, index laws etc. if you hope to successfully tackle the sort of calculus questions you're going to get given n the coming days and weeks. - Feb 22nd 2009, 05:20 AMTYTY