Integral of 10 / (5x^2 - 2x^3) Do not determine the numerical values of the coefficients.
Just wanted to know why when I write out the form of the partial fraction decomposition
there is A/x + B/x^3 + C/ (x^2 + 4)
Why isn't it just A/x^3 + B/ (x^2 + 4)?
your denominator is
5x^2 - 2x^3
= x^2(5-2x)
so you could have had 3 fractions,
A/x + B/x^2 + C/(5-2x)
that would have given you a LCD of x^2(5-2x)
or 5x^2 - 2x^3
I don't see how you got your choice or even the supposedly correct choice.
Sorry I typed out the wrong answer. Why would I have 3 fractions though? Because it's x^2?
yes, the option I gave you was
A/x + B/x^2 + C/(5-2x)
of course it could have been only 2 fractions with
B/x^2 + C/(5-2x) , and the LCD would also have been
x^2(5-2x)
but we don't know that, so we have to take the worst-case scenario.
To find the partial fraction decomposition of a rational function, you need to consider the degree of the denominators in the factors. In your case, the denominator is 5x^2 - 2x^3.
When you first factor the denominator, you get: 5x^2 - 2x^3 = -2x^3 + 5x^2.
Next, let's factor out x^2: -2x^3 + 5x^2 = x^2(-2x + 5).
Now, we can see that -2x + 5 is a linear factor, and x^2 is a quadratic factor.
When you perform the partial fraction decomposition, you decompose the rational function into simpler fractions based on the factors of the denominator.
For the linear factor -2x + 5, you write it as A/x, where A is an unknown constant.
For the quadratic factor x^2, you write it as B/(x^2 + 4), where B is another unknown constant. Notice that it is x^2 + 4, not just x^2. This is because it represents a quadratic factor that cannot be further factored, which means it stays intact in the decomposition.
If you were to write it as A/x^3 instead, you would miss out on the quadratic factor x^2, and the partial fraction decomposition would be incomplete.
In summary, when performing the partial fraction decomposition, you need to consider the degrees of the factors in the denominator and write the appropriate terms for each factor. That's why you have both A/x and C/(x^2 + 4) in the decomposition.