integral of 1/(9 + 4x^2)
The book says u should be 2/3x. Can someone enlighten me on why this is? I am dumbfounded
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
$\displaystyle 9 \cdot \left( \frac{2x}{3} \right)^2 = 9 \cdot \frac{2x}{3} \cdot \frac{2x}{3} = 9 \cdot \frac{4x^2}{9} = 4x^2$.
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.