Hey! I need help Evaluating the integral of [5*(x^3)-7x+4]/[(x^2)*(x-1)] dx using partial fractions...
It's traditional to use division first to reduce the numerator to some degree less than the degree of the denominator.
is this a long division problem, my original denominator was \[den=(x^{3}-x^{2})\] but that is linear, isn't it?
Both numerator and denominator are degree 3.
wouldn't I go straight over to partial fractions?
Your life will be better if you resist that temptation. Division will give you at least one trivial term for your integrand. Why would you not want that?
Long division gives \(5 + \dfrac{5x^{2} - 7x + 4}{x^2(x-1)}\). See how nice and easy that '5' out front looks?!
that 5 does look pretty nice...let me run through the long division to see if I have any problems on that
You sound converted. Sweet!!
that was definitely the best response I've seen all week
and I got the division correct, sweet!
Don't tell anyone, but we can actually have a little fun doing mathematics. Okay, now off to the Partial Fractions. What are your denominators?
\[x, x^{2},(x-1)\] I believe that should be the decomposition
You have this. Pop it out and let's see what you get.
Oh...that's why I flipped my biscuits...the way I would have done it would have had a cubic against a quadratic. I do believe I will get: \[\int\limits_{}^{}\frac{5x^2-7x+4}{x^3-x^2} dx=\int\limits_{}^{}\frac{3}{x}dx+\int\limits_{}^{}\frac{-4}{x^{2}}dx+\int\limits_{}^{}\frac{2}{(x-1)}dx\] which yields \[3\ln(\left|x\right|) + \frac{4}{x} + 2\ln(\left| x-1 \right|) + C\]
Except for one thing. Something else is entirely missing. What is it?
+5 xD
That's it. We put it out there to be all pretty. No sense forgetting about it. Excellent work.
I got a no-go from mathlab....are there any alternative forms
Perhaps I mistook your xD. That should be "5x".
that deserves at least three awkard turtles. Thank you for the help!
Just a tiny glitch. Now we have it.
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