Integration. How do I find the primitive
$$F(x)=\int \frac{1}{(x-2)^2}$$ ? Hints appreciated.

Question

Integration. How do I find the primitive 
$$F(x)=\int \frac{1}{(x-2)^2}$$ ? Hints appreciated.

raden · Answer

int by u-sub
let : u=x-2

anonymous · Answer

I would recommend you perform an u-substitution for this kind of problem:

$$\Large u=x-2 
\
\Large du=dx\ $$

terenzreignz · Answer

Before anything else, and while this may seem to be immaterial, I think it matters if you're to understand u-substitution method...
It's that your integral doesn't make sense if you don't indicate by what variable you integrate.  Better rewrite with the simple modification...
$$F(x)=\int \frac{1}{(x-2)^2}dx$$

terenzreignz · Answer

As they suggested, you now begin the u-substitution method.
You let u = x - 2
Then you get the derivative of u, with respect to x...
$$\large \frac{du}{dx} = 1$$
And you 'multiply both sides by dx'
$$\large du = 1\cdot dx = dx$$

terenzreignz · Answer

Returning to your integral, we make the necessary substitutions, namely, replacing x-2 with u, and replacing dx by du, as we derived.

$$\large \int\limits \frac{1}{u^2}du$$
And now you see the importance of the differential 'du' at the integral, it tells you that now, we integrate with respect to u.  Can you do it from here?

anonymous · Answer

Yes. $$\int u^{-2}du= -\frac{1}{u}+C. \text{Since u=(x-2)} ... -\frac{1}{x-2}+C$$ Worked like a charm

terenzreignz · Answer

Nicely done :)
I'd usually get rid of that ugly minus sign and make it
$$\frac{1}{2-x}+C$$
But that's fine :)