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Calc help

Blogs > PocketX
Post a Reply
PocketX
Profile Blog Joined January 2009
53 Posts
March 17 2009 22:16 GMT
#1
Evaluate:
[image loading]


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
Profile Blog Joined July 2006
United States24649 Posts
Last Edited: 2009-03-17 22:22:01
March 17 2009 22:21 GMT
#2
Only thing that makes sense is partial fractions to split this up.
ModeratorThere are animal crackers for people and there are people crackers for animals.
Nytefish
Profile Blog Joined December 2007
United Kingdom4282 Posts
March 17 2009 22:27 GMT
#3
It splits nicely into 3 terms.
No I'm never serious.
Divinek
Profile Blog Joined November 2006
Canada4045 Posts
Last Edited: 2009-03-17 22:33:39
March 17 2009 22:30 GMT
#4
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.
Never attribute to malice that which can be adequately explained by stupidity.
Oh goodness me, FOX tv where do you get your sight? Can't you keep track, the puck is black. That's why the ice is white.
naonao
Profile Blog Joined November 2008
United States847 Posts
March 17 2009 22:32 GMT
#5
split it into 1/x^2 - 1/x - 1/(x+1) then integrate
DeathByMonkeys
Profile Blog Joined March 2008
United States742 Posts
March 17 2009 22:34 GMT
#6
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
Profile Blog Joined January 2009
53 Posts
March 17 2009 22:35 GMT
#7
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
Profile Blog Joined November 2008
United States847 Posts
March 17 2009 22:40 GMT
#8
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
Profile Blog Joined April 2007
United States1133 Posts
March 17 2009 22:40 GMT
#9
Yeah, 1 is a polynomial. A 0th degree polynomial, but a polynomial nonetheless.
skating
DeathByMonkeys
Profile Blog Joined March 2008
United States742 Posts
Last Edited: 2009-03-17 22:42:37
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
Profile Blog Joined January 2009
53 Posts
March 17 2009 22:43 GMT
#11
Thanks everyone I got it.
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