Question

I have a question.

Answer 1

hint:

2x + 1 is linear and (x-2)^2 is repeated linear factor so your partial fraction will look like \[\frac{A}{2x + 1} + \frac{B}{x-2} + \frac C{(x-2)^2}\]
does that help?

Answer 2

Yeah it helped. But can you give me idea for other possible situation?

Answer 3

you mean for the second part?

Answer 4

or are you asking for an alternative solution to this?

Answer 5

not for a second part. how to find partial fractions of other questions of same type.

Answer 6

@curiousshubham could you clarify that? You want a general over view on how to do partial fraction expansion?

Answer 7

there are certain rules...
if the denominator is linear like for example \[\frac{1}{x-2}\]
the denominator is x- 2 which is linear.. the partial fraction would be \[\frac{A}{x-2}\]
or if you have \[\frac 2x\] partial fraction would be \[\frac{A}{x}\]
okay?

this is case I. linear denominators

Answer 8

yeah thats the thing i have been asking for.....

Answer 9

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

Answer 10

...so do i still have to go on a lengthy explanation about the cases?

Answer 11

lol. probably.

Answer 12

Thanks @Algebraic! for the link.
Thanks @lgbasallote . Nope.