<s>[INST] Evaluate the Integral: $\int\frac{5x+1}{(2x+1)(x-1)}dx$ $$\int\frac{5x+1}{(2x+1)(x-1)}dx$$
$$\frac{A}{(2x+1)}+\frac{B}{(x-1)}$$
$$\frac{A(x-1)+B(2x-1)}{(2x+1)(x-1)}$$
I put $x=1$ to attain B: $5\ (1)+1=A(1-1)+B\ (2(1)+1)$
$$6=0+3B$$
$$B=2$$
However, I am unable to utilize the same method to attain A.
What should I do? 
 [/INST] You had a mistake in your computations. See the first line below. You have written $B(2x-1)$ instead of $B(2x+1)$.
Consider the following steps
$$\eqalign{
  & {{5x + 1} \over {(2x + 1)(x - 1)}} = {A \over {2x + 1}} + {B \over {x - 1}} = {{A\left( {x - 1} \right) + B\left( {2x + 1} \right)} \over {(2x + 1)(x - 1)}}  \cr 
  & 5x + 1 = A\left( {x - 1} \right) + B\left( {2x + 1} \right)  \cr 
  & x = 1\,\,\, \to \,\,\,6 = 3B\,\,\, \to \,\,\,B = 2  \cr 
  & x =  - {1 \over 2}\,\,\,\, \to \,\,\, - {3 \over 2} =  - {3 \over 2}A\,\,\, \to \,\,\,A = 1  \cr 
  & {{5x + 1} \over {(2x + 1)(x - 1)}} = {1 \over {2x + 1}} + {2 \over {x - 1}} \cr} $$
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