Integrating partial fractions I have
$\int{\frac{2x+1}{x^2+4x+4}}dx$
Factorising the denominator I have
$\int{\frac{2x+1}{(x+2)(x+2)}}dx$
From there I split the top term into two parts to make it easier to integrate
$\int{\frac{2x+1}{(x+2)(x+2)}}dx$ = $\int{\frac{A}{(x+2)}+\frac{B}{(x+2)}}dx$
=$\int{\frac{A(x+2)}{(x+2)^2}+\frac{B(x+2)}{(x+2)^2}}dx$
Therefore
$2x+1 = A(x+2) +B(x+2)$
This is where I would normally use a substitution method to eliminate either the A term or B term by letting x = something like -2, which would get rid of the A and usually leave me with the B term to solve. However since they are the same I'm not sure what to do. 
I've been told to try evaluate the co-efficients, but am not sure how. 

All you need to do is to solve this with respect to polynomials: 
$2x+1=Ax+2A+Bx+2B$
$2x+1=x(A+B)+(2A+2B)$
$A+B=2 \rightarrow  B=2-A$
$2A+2B=1$
$2A+4-2A=1\rightarrow 4=1$
This is contradiction! You have made an mistake in step where you split the term into two fractions, you should have done it like this: 
$\frac{2x+1}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}$ 
and then proceed like usual.
$A(x+2)+B=2x+1$
$A=2$
$2A+B=1$
$B=-3$
$\int \frac{2}{x+2}+\frac{-3}{(x+2)^2}dx=2ln(x+2)+\frac{3}{x+2}+C$
