What is the solution to this basic integral using substitution?

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Homework Statement


Intergrate:

\int\frac{3 dx}{\left(2-x\right)^{2}}
By substituion.

Homework Equations



n/a

The Attempt at a Solution



Ok so first I take the integer out to get:

3\cdot\int\frac{dx}{\left(2-x\right)^{2}}

Now I let u = 2 - x and du = dx to get:

3\cdot\int\frac{du}{u^{2}}

Now I take away the fraction:

3\cdot\int u^{-2} du

Now I am stuck at this point. Any help would be apppreciated.
 
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This is a basic integral, with this I mean that such integral should be learned by heart.
You have that (if n\neq -1 )

\int{x^ndx}=\frac{1}{n+1}x^{n+1}+C

Just apply this formula with n=-2.
 
well, pretend that u^-2 is your x.

Do you know how to integrate x^-2?
(u⁻² ⁺¹)/(-2+1) + c is what you get.
don't forget that you still have that whole integral multiplied by 3.

When you're all finished, just substitute your 2-x back in and you've done the integral!:smile:
 
micromass said:
This is a basic integral, with this I mean that such integral should be learned by heart.
You have that (if n\neq -1 )

\int{x^ndx}=\frac{1}{n+1}x^{n+1}+C

Just apply this formula with n=-2.

I just found a sheet with a few calc problems at my tutoring job and its been months since I did any calc so I decided to brush up a bit and test how much i remembered. Obviously not much all things considered but you've refreshed my memory. Thanks :)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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