Integral of $\int \frac{5x+1}{(2x+1)(x-1)}$ I am suppose to use partial fractions
$$\int \frac{5x+1}{(2x+1)(x-1)}$$
So I think I am suppose to split the top and the bottom. (x-1)
$$\int \frac{A}{(2x+1)}+ \frac{B}{x-1}$$ 
Now I am not sure what to do.

Bring the expression $\frac{A}{2x+1}+\frac{B}{x-1}$ to the common denominator $(2x+1)(x-1)$. We get
$$\frac{A(x-1)+B(2x+1)}{(2x+1)(x-1)}.$$
The top is $(A+2B)x -A+B$.  We want this to be identically equal to $5x+1$. 
The coefficients of $x$ must match, and the constant terms must match.
So we want $A+2B=5$, and $-A+B=1$. Solve for $A$ and $B$. We get $B=2$ and $A=1$.
Check by cross-multiplying that we did not make a mistake. 
Now find 
$$\int\left(\frac{1}{2x+1}+\frac{2}{x-1}\right)\,dx.$$
Remark: Another popular technique is to stop at $A(x-1)+B(2x+1)=5x+1$. Plug in $x=1$. We get $3B=6$, so $B=2$. Plug in $x=-1/2$ (to kill the $2x+1$ term). We get $(-3/2)A=-3/2$, so $A=1$.
