Calc help

PocketX

53 Posts

March 17 2009 22:16 GMT

Evaluate:

How to solve this? I'm a first year non-math university student.
The only approach that seems we can apply here is either substitution or integration by parts: uv - /v du where / is the integration sign.

I've tried substitution and it would go in circles, and by parts I always create a ln in the integration part and it seems to complicated to continue from there.

This question should be a regular and not challenging question, but I can't solve it. Help plz!

micronesia

United States24768 Posts

March 17 2009 22:21 GMT

Only thing that makes sense is partial fractions to split this up.

Nytefish

United Kingdom4282 Posts

March 17 2009 22:27 GMT

It splits nicely into 3 terms.

Divinek

Canada4045 Posts

March 17 2009 22:30 GMT

http://tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx

Or if you're that lazy pull out the common x so you have 1/x^2(x+1) and then your partial fraction is

A/x + B/x^2 + C/x+1 = 1

solve for the three then re sub them in.

naonao

United States847 Posts

March 17 2009 22:32 GMT

split it into 1/x^2 - 1/x - 1/(x+1) then integrate

DeathByMonkeys

United States742 Posts

March 17 2009 22:34 GMT

Like the others said split it into two fraction then integrate...

after you split it into two the two fractions will also equal:

x^-3 + x^-2

then it'll be easier to integrate

PocketX

53 Posts

March 17 2009 22:35 GMT

Thanks everyone, I also learnt partial fractions, so I know how to do it, but iirc, partial fractions is polynomials on the numerator and denominator as in f(x)/g(x) for it too work. But in this case, is the numerator a polynomial?

naonao

United States847 Posts

March 17 2009 22:40 GMT

yes, constant values are also polynomials, they just must have all terms in the form ax^n where n is a nonnegative integer and a is real

huameng

United States1133 Posts

March 17 2009 22:40 GMT

Yeah, 1 is a polynomial. A 0th degree polynomial, but a polynomial nonetheless.

DeathByMonkeys

United States742 Posts

March 17 2009 22:42 GMT

#10

On March 18 2009 07:35 PocketX wrote:
Thanks everyone, I also learnt partial fractions, so I know how to do it, but iirc, partial fractions is polynomials on the numerator and denominator as in f(x)/g(x) for it too work. But in this case, is the numerator a polynomial?

Ok so basically you split it into

int((1/x^3) + (1/x^2))

1/x^3 = x^ -3
1/x^2 = x^ -2

integrating x^ -3 = (x^-2) / -2
integrating x^ -2 = (x^ -1) / -1 = -(x^ -1)

EDIT: Shit this is so much quicker and easier to read on paper

PocketX

53 Posts

March 17 2009 22:43 GMT

#11

Thanks everyone I got it.

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