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.

But $$\frac{5x+1}{(2x+1)(x-1)}\ne\frac{5x+1}{2x+1}+\frac{5x+1}{x-1}\;,$$ as you’ll see if you combine the fractions on the righthand side over a common denominator: you get
$$\begin{align*}
\frac{5x+1}{2x+1}+\frac{5x+1}{x-1}&=\frac{5x+1}{2x+1}\cdot\frac{x-1}{x-1}+\frac{5x+1}{x-1}\cdot\frac{2x+1}{2x+1}\\\\
&=\frac{(5x+1)(x-1)+(5x+1)(2x+1)}{(2x+1)(x-1)}\\\\
&=\frac{(5x+1)\big((x-1)+(2x+1)\big)}{(2x+1)(x-1)}\\\\
&=\frac{(5x+1)(3x)}{(2x+1)(x-1)}\;,
\end{align*}$$
clearly not the same as $$\frac{5x+1}{(2x+1)(x-1)}\;.\tag{1}$$ You have to do a bit of algebra to split into partial fractions. Set it up in the usual way:
$$\frac{5x+1}{(2x+1)(x-1)}=\frac{A}{2x+1}+\frac{B}{x-1}=\frac{A(x-1)+B(2x+1)}{(2x+1)(x-1)}\tag{2}\;.$$
We want to choose $A$ and $B$ so that the fractions on the two ends of $(2)$ really are equal; they have the same denominator, so they must have the same numerator, and therefore $$A(x-1)+B(2x+1)=5x+1$$ or, after multiplying out and collecting terms on the lefthand side, $$(A+2B)x+(-A+B)=5x+1\;.\tag{3}$$ The only way that the two sides of $(3)$ can be the same polynomial is to have $A+2B=5$ and $-A+B=1$. Solve this little system for $A$ and $B$, which turn out to be nice numbers, and plug those values back into the middle expression in $(2)$. You’ll end up with the integration $$\int\frac{A}{2x+1}dx+\int\frac{B}{x-1}dx$$ (with specific numbers for $A$ and $B$), and this is pretty straightforward.
