partial fraction decompositions

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January 27th 2008, 11:56 AM
jmaillou

partial fraction decompositions

So I have an assignment, and we are supposed to integrate using partial fraction decompositions. The integral is:

(x^2 + 53)/(x^3 + x^2)

What I have so far is:
A/(x) + (Bx+C)/(x^2) + D/(x +1)

Then:
A(x^2 + x) + (Bx + C)(x+1) + D(x^2) = (x^2 + 53)

I've determined that C= 53, and D= 54, but I'm stuck finding A & B. Any help would be much appreciated. Once I figure out what these partial fractions are, I'll be home free, since integrating is the easy part.

Thanks :)
January 27th 2008, 12:11 PM
Jhevon

Quote:

Originally Posted by jmaillou

So I have an assignment, and we are supposed to integrate using partial fraction decompositions. The integral is:

(x^2 + 53)/(x^3 + x^2)

What I have so far is:
A/(x) + (Bx+C)/(x^2) + D/(x +1)

Then:
A(x^2 + x) + (Bx + C)(x+1) + D(x^2) = (x^2 + 53)

I've determined that C= 53, and D= 54, but I'm stuck finding A & B. Any help would be much appreciated. Once I figure out what these partial fractions are, I'll be home free, since integrating is the easy part.

Thanks :)

one way: expand the left hand side:

so we have: $(A + B + D)x^2 + (A + B + C)x + C = x^2 + 53$

now equate the coefficients of like powers of x.

from this we see:

$A + B + D = 1$ ..............(1)

$A + B + C = 0$ ...............(2)

$C = 53$ .........................(3)

since you know C and D, solving for A and B should be easy from this system ...oh wait, we still have a clue missing

there might be some virtue in doing $\frac {x^2}{x^2 (x + 1)} + \frac {53}{x^2(x + 1)}$

so the partial fractions are: $\frac 1{x + 1} + \frac {53}{x^2(x + 1)}$

if you desire you can work on the second fraction a bit more to see if you can get partial fractions from it
January 27th 2008, 12:16 PM
Peritus

I think maybe the problem you had is that you have only 3 equations and 4 unknowns, and after finding D and C you get two identical equations:

A+B = -53

------------------------------
so you can choose the value of either A or B as you wish, for example a convinient choice would be B=0 and thus A=-53...
January 27th 2008, 12:19 PM
Jhevon

Quote:

Originally Posted by Peritus

I think maybe the problem you had is that you have only 3 equations and 4 unknowns, and after finding D and C you get two identical equations:

A+B = -53

------------------------------
so you can choose the value of either A or B as you wish, for example a convinient choice would be B=0 and thus A=-53...

yes, i saw that afterwards. i edited my post. don't know how much better i made the situation. too lazy to check it