Simple partial fractions questions with fractions in the numerator.

Question

$$ \dfrac{1}{\left(x+1\right)\left(2x+1\right)}$$

Using the cover up rule, I received the answer of $$\dfrac{\dfrac{1}{2}}{2x+1}-\frac{1}{x+1}$$

or

$$\dfrac{1}{2(2x+1)}-\frac{1}{x+1}$$

Answer at the back of the book says:

$$\dfrac{2}{2x+1}-\frac{1}{x+1}$$

I may have a gap in my knowledge here, how did the book receive that answer?

Answer 1

We have $$\frac{1}{(2x+1)(x+1)} \equiv \frac{A}{2x+1} + \frac{B}{x+1}$$

So, multiply through by $(2x+1)(x+1)$ to get $$1 = A(x+1) + B(2x+1)$$

Let $x = -1$ to get $1 = B \cdot -1 \iff B = -1$

Let $x = -\frac{1}{2}$ to get $\frac{A}{2} = 1 \iff A = 2$.

So, replacing your values of $A$ and $B$ into the first equation we get:

$$\bbox[8px, border:2px lightblue solid]{\frac{1}{(2x+1)(x+1)} \equiv \frac{2}{2x+1} - \frac{1}{x+1}}$$

Answer 2

Suppose $\displaystyle \frac{1}{(x+1)(2x+1)}=\frac{A}{x+1}+\frac{B}{2x+1}$.

Then by equating the numerators, $1=2Ax+A+Bx+B$.

Now comparing coefficients, $A=-1$ and $B=2$.