# Trying to figure out integral

Hi,

I have a couple of problems I'm trying to figure out:

1) 1/x^3-x^2 dx

Is this a partial fraction problem and if so am I doing it correctly?

I think I should break down the denominator into x^2(x-1) then I have A/x^2 + B/(x-1)

Then I get A= -3 and B = 1 so I figure it to be -3ln[x^2] + ln[x-1]?

I may be way off and if I am could you please point me in the right direction.

2) [5x^3 + x^2 - square root (x)] / 2x

I know this is another partial fractions problem but I have no idea what to do when I only have 2x in the denominator. Could you give me a way to start this problem so I can try to figure it out too?

Thanks for the help

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Char. Limit
Gold Member

Hi,

I have a couple of problems I'm trying to figure out:

1) 1/x^3-x^2 dx

Is this a partial fraction problem and if so am I doing it correctly?

I think I should break down the denominator into x^2(x-1) then I have A/x^2 + B/(x-1)

Then I get A= -3 and B = 1 so I figure it to be -3ln[x^2] + ln[x-1]?

I may be way off and if I am could you please point me in the right direction.
Okay, here we go. First, you're right that it's a partial fractions problem. However, when you have x*x*(x-1), you have to have a term for 1/x^2 and 1/x. So in reality, you have A/x^2 + B/x + C/(x-1). That will help.

Also, something that will help is that the integral of 1/x^2 is not ln(x^2). You have to use the power rule for that one.

2) [5x^3 + x^2 - square root (x)] / 2x

I know this is another partial fractions problem but I have no idea what to do when I only have 2x in the denominator. Could you give me a way to start this problem so I can try to figure it out too?

Thanks for the help
This one is actually really easy. Just split the numerator and then you'll get a formula in terms of x^2, x, and 1/sqrt(x). Each of these can be integrated using the power rule.

HallsofIvy
Homework Helper

Actually, it's impossible to tell if it is a "partial fractions" problem without knowing whether you meant
$$\frac{1}{x^3}- x^2$$
(what you actually wrote)
or
1/(x^3- x^2) which would be
$$\frac{1}{x^3- x^2}$$

If it is the latter, then, yes, factor and write as partial fractions.