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.
 A: Suppose $\frac{5x+1}{(2x+1)(x-1)} = \frac{A}{2x+1} + \frac{B}{x-1}$. Then $5x+1 = A(x-1)+B(2x+1) = (A+2B)x + (B-A)$. So by comparing coefficients, we get $A+2B = 5$ and $B-A = 1$. Solving this gives $A = 1$ and $B = 2$.
A: 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 $A=1$.
A: 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.
A: From Concrete Mathematics section 7.3 SOLVING RECURRENCES page 340, edited.

If $R(z) = P(z)/Q(z)$, where $Q(z) = q_0(1-\rho_1z)\ldots(1-\rho_lz)$ and the numbers $(\rho_1,\ldots,\rho_l)$ are distinct, and if $P(z)$ is a polynomial of degree less than $l$, then
  \[
  R(z) = \frac{a_1}{1-\rho_1z} + \cdots + \frac{a_l}{1-\rho_lz},\qquad\hbox{where $a_k = \frac{-\rho_kP(1/\rho_k)}{Q'(1/\rho_k)}$}
  \]
  Proof: Let $a_1,\ldots,a_l$ be the stated constants, and
  \[
  S(z) = \frac{a_1}{1-\rho_1z} + \cdots + \frac{a_l}{1-\rho_lz}
  \]
  And can prove that $R(z) = S(z)$ by showing that the function $T(z) = R(z) - S(z)$ is not infinite as $z \to \infty$; hence $T(z)$ must be zero.
Let $\alpha_k = 1/\rho_k$. To prove that $\lim_{z\to\alpha_k}T(z)\neq\infty$, it suffices to show that $\lim_{z\to\alpha_k}(z-\alpha_k)T(z)=0$, because $T(z)$ is a rational function of $z$. Thus we want to show that
  \[
  \lim_{z\to\alpha_k}(z-\alpha_k)R(z) = \lim_{z\to\alpha_k}(z-\alpha_k)S(z)
  \]
  The right-hand limit equals $\lim_{z\to\alpha_k}a_k(z-\alpha_k)/(1-\rho_kz)=-a_k/\rho_k$, because $(1-\rho_kz) = -\rho_k(z-\alpha_k)$ and $(z-\alpha_k)/(1-\rho_jz) \to 0$ for $j \neq k$. The left-hand limit is
  \[
  \lim_{z\to\alpha_k} (z-\alpha_k)\frac{P(z)}{Q(z)} = P(\alpha_k)\lim_{k\to\alpha_k}\frac{z-\alpha_k}{Q(z)} = \frac{P(\alpha_k)}{Q'(\alpha_k)}
  \]
  by L'Hospital's rule. Thus the theorem is proved.

