Integrating through partial fractions

[ACLerok]

here's the problem, i am supposed to take the integral from 1 to 2 of this:

(dx)/[(X+3)^2 (x+1)^2]

I decided that the easiest way to compute it is by integrating by partial fractions so what i did was set up the equation:

A/(x+3) + B/[(x+3)^2] + C/(x+1) + D/[(X+1)^2] = 1

After this I multiplied by the common denominator and got:

[A(x+3)(x+1)^2] + [B(x+1)^2] + [C(x+1)(x+3)^2] + [D(x+3)^2] = 1

This is where I am stuck. I can get the values of coefficients B and D by pluggin in x=-3 and x=-1 respectively by I don't know how to find coefficients A and B. I tried multiplying everything out but that just made things more complicated and confusing.

Anyone kind enough to help me out?
Thanks!

 

[arildno]

Multiplying out works just fine, as long as you group your terms correctly:
Here's the coefficients for each power of x:
$$x^{3}:A+C$$
$$x^{2}:5A+B+7C+D$$
$$x^{1}:7A+2B+15C+6D$$
$$x^{0}:3A+B+9C+9D$$
In order for this polynomial to be identically equal to 1, we must solve the system:
A+C=0
5A+B+7C+D=0
7A+2B+15C+6D=0
3A+B+9C+9D=1

You ought to try again and group the terms in this manner.
Good luck!

 

[ACLerok]

i don't see how you can solve these questions with just one equation with two variables and the others with all four. Unless i am missing something...

Multiplying out works just fine, as long as you group your terms correctly:
Here's the coefficients for each power of x:
$$x^{3}:A+C$$
$$x^{2}:5A+B+7C+D$$
$$x^{1}:7A+2B+15C+6D$$
$$x^{0}:3A+B+9C+9D$$
In order for this polynomial to be identically equal to 1, we must solve the system:
A+C=0
5A+B+7C+D=0
7A+2B+15C+6D=0
3A+B+9C+9D=1

You ought to try again and group the terms in this manner.
Good luck!

 

[arildno]

You have 4 equations in 4 unknows, right?
Remember, your own equation:
[A(x+3)(x+1)^2] + [B(x+1)^2] + [C(x+1)(x+3)^2] + [D(x+3)^2] = 1

This must hold for ALL choices of x!
This is only possible if the coefficients before $$x^{3},x^{2},x$$ are all ZERO!
The constant term must equal 1.

 

[Parth Dave]

It seems like a huge pain solving a system of four equations. However, sometimes what you can do (in this question for example) is plug in your own choices for x and then watch things cancel out. For example,

[A(x+3)(x+1)^2] + [B(x+1)^2] + [C(x+1)(x+3)^2] + [D(x+3)^2] = 1

If x = -3 than, A, C, D terms cancel out and you are left with
B(-3 + 1)^2 = 1. Thus, B = 1/4
You can do the same thing with x = -1. Now you only have a two variables to solve for. Not all that difficult or time consuming.

 

[arildno]

That's what he started with Parth Dave; he wanted another method..

 

[Parth Dave]

Oh, uh... my mistake... next time I'll read everything .

 

[maverick280857]

It seems like a huge pain solving a system of four equations. However, sometimes what you can do (in this question for example) is plug in your own choices for x and then watch things cancel out. For example,

[A(x+3)(x+1)^2] + [B(x+1)^2] + [C(x+1)(x+3)^2] + [D(x+3)^2] = 1

If x = -3 than, A, C, D terms cancel out and you are left with
B(-3 + 1)^2 = 1. Thus, B = 1/4
You can do the same thing with x = -1. Now you only have a two variables to solve for. Not all that difficult or time consuming.

HI. I suppose you've done this but one way to proceed further from Parth Dave's equation above is to realize that it is an identity for the quadruplet (A,B,C,D) that is to be found. Hence, you can differentiate both sides successively with respect to x. That will give you lower order algebraic equations that can be easily solved instead of comparing coefficients right from the start.

Cheers
Vivek

 

[ACLerok]

In solving for A and C, can i plug in the vaules of B and D in the system of equations to help solve for A and C?

 

[Tide]

Why not do your partial fraction decomposition in two stages? E.g.

$$\left( \frac {1}{(x+1)(x+3)} \right)^2 = \frac {1}{4} \left( \frac {1}{x+1} - \frac {1}{x+3} \right)^2$$

You can expand the square and do another partial fraction decomposition for the middle term.