| Blogs > PocketX |
|
PocketX
53 Posts
![[image loading]](http://img25.imageshack.us/img25/6830/123duj.jpg) 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 States24805 Posts
| ||
|
Nytefish
United Kingdom4282 Posts
| ||
|
Divinek
Canada4045 Posts
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
| ||
|
DeathByMonkeys
United States742 Posts
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
| ||
|
naonao
United States847 Posts
| ||
|
huameng
United States1133 Posts
| ||
|
DeathByMonkeys
United States742 Posts
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
| ||
| ||
|
|
AI Arena Tournament
BSL: Ladder Tournament
Sparkling Tuna Cup
Patches Events
BSL Open Qualifier
BSL Open Qualifier
OSC
OSC
WardiTV Weekly
PiGosaur Cup
[ Show More ] |
|
|
EPT
