Maths integration help needed!

I've forgotten how to integrate functions in the form of a fraction where the integral is not a log natural... can someone help me with this question?

∫ 2 / (x-2)^2

change it into the form 2(x-2)^-2

then divide by the new power which would be -1, since the coefficient of x is 1 it has no effect

dividing 2 by -1 gives -2

so it becomes -2(x-2)^-1

edit: you can also test the answer by differentiating it, so -1 x -2 = 2

so therefore 2 x (x-2)^-2 = 2(x-2)^-2 which is the original function

Rewrite that as $\displaystyle 2 \int \frac{\mathrm{d}x}{(x-2)^2} = 2 \int (x-2)^{-2} \, \mathrm{d}x$ using the fact that $\frac{1}{a^b} = a^{-b}$.

Then use the fact that $\displaystyle \int x^n \, \mathrm{d}x = \frac{x^{n+1}}{n+1} + \mathcal{C}$, what does that get you?

Rewrite that as $\displaystyle 2 \int \frac{\mathrm{d}x}{(x-2)^2} = 2 \int (x-2)^{-2} \, \mathrm{d}x$ using the fact that $\frac{1}{a^b} = a^{-b}$.

Then use the fact that $\displaystyle \int x^n \, \mathrm{d}x = \frac{x^{n+1}}{n+1} + \mathcal{C}$, what does that get you?

what about this one? ∫ x / (4 - x^2)^1/2 dx

thank you once again

Use the substitution $u = 4-x^2$.

Surely inspection is the easiest method here.

That's the way I'd do it as well, but he didn't spot the inspection so I suggested the sub. Which is fairly equivalent, really.

In an exam it's best to do both methods to double check, you don't want to be losing marks when it comes to something as trivial as integration