Are these two partial fraction the same?

Question

I tried solving a pretty standard partial fraction equation. $$ \frac{5-x}{2x^2+x-1} $$

Which becomes:

$$ \frac{A}{x-0.5} + \frac{B}{x+1} $$

Solving the partial fraction: $$ 5-x = A(x+1)+B(x-0.5) $$ $$ 5-x = Ax+A+Bx-0.5B $$ $$ 5 = A -0.5B $$ $$ -x = Ax +Bx $$ Solved simultaneous: $$ B = -4, A = 3 $$

However, the textbook proposed to first multiply the denominator with the half. $$ \frac{A}{2x-1} + \frac{B}{x+1} $$ Which gives: $$ 5-x = Ax+A+2Bx-B $$ And the final answer of: $$ B = -2, A =3 $$

Are these two solutions the same?

Answer 1

I'm sorry. I was a little two quick. If we put $A=3,B=-4$ back in to the fractions and you get

$$ \frac{A}{x-0.5} + \frac{B}{x+1}= \frac{3}{x-0.5} - \frac{4}{x+1} = \frac{6}{2x-1} - \frac{4}{x+1} = {6(x+1)-4(2x-1)\over (2x-1)(x+1)}= {-2x+10\over (2x-1)(x+1)} $$

So you have to be careful. Take a look at Tony's answer.

Answer 2

You've got a problem in your top derivation. If it's true that $$\frac{5-x}{2x^2+x-1} = \frac{A}{x-\frac12}+\frac{B}{x+1},$$

then it doesn't follow that

$$5-x = A(x+1) + B\left(x-\frac12\right)$$

You've lost a factor of $2$. If you multiply both sides of the initial equation by a common denominator, you get $$5-x=2A(x+1)+2B\left(x-\frac12\right)$$

Answer 3

Since you chose to decompose $$\frac{5-x}{2x^2+x-1}$$ as $$\frac{A}{x-0.5}+\frac{B}{x+1}$$ you're forgetting that $$2x^2+x-1=[2(x-0.5)](x+1)$$ So the statement $$5-x=A(x+1)+B(x-0.5)$$ is not true. The correction should be $$5-x=2A(x+1)+2B(x-0.5)$$ which eventually results in $A=\frac{3}{2}$ and $B=-2$. Again, these are not the same as your textbook, but they will have the same result when integrating. This is because your textbook chooses the decomposition $$\frac{5-x}{2x^2+x-1}=\frac{A'}{2x-1}+\frac{B}{x+1}$$ Now, take in count that this has to be true: $$\frac{A}{x-0.5}=\frac{A'}{2x-1}$$ This can only mean that $A'=2A$. So that's why your choice should give you a result for $A$ that is half the answer from your textbook. It doesn't matter the choice you take to decompose the fraction, as long as the procedure is done correctly, the integrated value will be the same.