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

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?

Zain Patel
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Modrisco
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    At one point you probably had $A(x+1)+B(2x+1)=1$. When you put $x=-1/2$ you should get $A(1/2)=1$. You probably mistakenly concluded $A=1/2$. It should be $A=2$. – André Nicolas Jul 05 '15 at 15:03

2 Answers2

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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:

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

Zain Patel
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  • is there a rule that says I have to definitely finish the solution before posting?Just for my own learning really. – John_dydx Jul 05 '15 at 15:07
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    Not particularly (or at least I don't know of any), but you need to be ready for the barrage of downvotes that come upon you as you post since you have an incomplete answer. I'd recommend posting with something substantial and then completing your answer. – Zain Patel Jul 05 '15 at 15:09
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    Thanks-I see the reason for all the downvotes now! – John_dydx Jul 05 '15 at 15:10
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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 $B=2$.

Extremal
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