Partial Fractions question (1 Viewer)

[NewiJapper]

Partial fractions is one of the only techniques of integration I'm still coming to terms with, any help with this one would be appreciated

 

[stevey6404]

Let that be

A/ (x-2) + B/(x-2)^2 + C (x-3)

and then do the same thing to find A, B and C and bam solve.

 

[NewiJapper]

I did that and got A+B+C=3 and 3A+2B+C=4 but then got a messy with the simultaneous equation haha

 

[kaz1]

equate coefficients of x² first

 

[schroot_92]

If you get to (x^2+3x+4)/((x-2)^2(x-3)) = A/((x-2)^2) + B/(x-2) + C/(x-3), then you can say that (x^2 + 3x + 4) = A(x-3) + B(x-2) + C(x-2)^2.. then if you let x=2, the B and C terms will go to 0, allowing you to easily find A (14 = -A).. Once you have found A, you can expand the right hand side, collect like terms and equate the coefficients, starting with the x^2 terms to find C, and then subbing in A and C to find B.

 

[funnytomato]

There are about 3 ways of finding the constants(limit, equating coefficients, substitute values), and it's up to you to choose one or a combination of these that's most convenient.

For this, let LHS= A/(x-2)^2 + B/(x-2) + C/(x-3)

A and C can be found using limit. You muptiply both sides by the denom. under whichever constant you're looking for
For A, muptiply both sides by (x-2)^2, so every thing on RHS but A becomes 0 as x approaches 2. And we can find A by evaluating the LHS as limit x ---> 2.
So A= lim x--> 2 ( x^2+3x+4)* (x-2)^2 / (x-3)*(x-2)^2= 4+6+4/-1 = -14 (sorry I failed to learn Latex )

similarly C=22

Then you equate coeff. of x^2 on both sides after you multiply both sides by (x-3)*(x-2)^2

we have 1 = B + C , B=1-C= -21