Hello Readers!

Could it be true? An easier way to do partial fractions? This has to be the most tedious thing that you learn in all of Calculus. Well, I’m here to tell you that you should dread it no longer! No more boring systems of equations!

It is too easy to make a mistake with the traditional method of partial fractions. Although solving a system of equations is a topic covered way back in algebra, nobody ever truly masters it. Why is that? Because you fall asleep half way through the problem! You end up not paying attention fully, and making a simple error somewhere along the way. You are probably thinking, “Mike, just tell us already!” So I will!

Conventional partial fractions are done as follows:

Ughhh… that was so tedious!! Please let me show you the faster way now! Let’s look at the same problem from the beginning.

At this point, let’s take a step back and think about the equation in front of us. The constants, A, B & C, have to make this true for ANY value of x. x could be absolutely anything, and this will always be true. So, why not choose the values of x? Pick the ones that would make this problem a lot easier for us.

How about if x = 2? That would cancel out one of the terms, and simplify things quite a bit.

Now, what if we let x = -3? That would also cancel out one of the terms, right?

And finally, what if we let x = 0? That would cancel out the Ax.

As you can see, we get the exact same result!! Isn’t that so exciting? We completely avoided the grueling part of partial fractions! There is so much less room to make errors, and it is much faster! Please try it out! I’ll post one more example below.

See! Isn’t that much, much easier? Well, I hope you can try it out. I know that partial fractions do not come up all that often, but when they do, you will be very glad you learned this method!

If you liked this post, I bet you would really love my U-Substitution shortcut.

Please comment below or send me an email if you have any questions at all!

## 11 comments

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## ktscopes says:

June 17, 2012 at 12:17 am (UTC -5)

You’re welcome!

## ktscopes says:

April 29, 2012 at 5:17 pm (UTC -5)

Excellent explanation, but the method works even for finding A. Let x = 2i and make the substitutions. You get

20i = (2A + 2)(2i + 4)(2i + 1)

20i = (2A + 2)(10i)

2 = 2Ai + 2

A = 0

If you let x = -2i you also arrive at A = 0

## mandy says:

April 7, 2012 at 8:33 am (UTC -5)

thnku so much..this made my complex integration easier for my exam prep..thnku :)

## Gavin Francis says:

April 4, 2012 at 10:19 pm (UTC -5)

thanx a lot !!!!!!!!!!!!!!!!

it’s surely going to help me while doing laplace transforms.

well…………do you have some shortcut tricks to find inverse of laplce transforms ????

## Mohammad says:

December 5, 2011 at 2:43 am (UTC -5)

cool,

I had already seen it though in a book. but had forgotten and sold the book, so this was as good as the first one(even better).

Thx Mike ,keep up the good work.

## Mike says:

October 5, 2011 at 10:09 pm (UTC -5)

The truth is that this DOES work for these types of problems, if you set them up correctly. The reason that it fails for this problem is because of the (s^2 + 4) in the denominator. The trick we use to avoid doing partial fractions the hard way, is that we let s equal a number that will make this term be zero. There is NO number that makes s^2 + 4 = 0. So, there is no number we can substitute in. Therefore, we can use this trick to solve the rest of the problem, but resort to the regular method for finding the last constant. Does that make sense? I’ll show you.

10s/((s^2 +4)(s+4)(s+1)) = (As + B)/(s^2 + 4) + C/(s+4) + D/(s+1)

10s = (As + B)(s + 4)(s + 1) + C(s^2 + 4)(s + 1) + D(s^2 +4)(s+4)

Let s = -1. (To get rid of all terms besides D)

-10 = D(5)(3)

-10 = 15D

D = -2/3

Let s = -4. (To get rid of all terms besides C)

-40 = C(20)(-3)

-40 = -60C

C = 2/3

Let s = 0. (To get rid of the term A, and use C and D to find B)

0 = 4B + 4C + 16D

B = -C -4D

B = -(2/3) – 4(-2/3)

B = -2/3 + 8/3

B = 6/3

B = 2

Now, we would try to make (s^2 + 4) = 0, but we can’t!!!!!!!

So, what do we do instead?

The answer is that you have to resort to the original method of partial fractions. This is a rare case, but it happens. This is the only reason this method doesn’t completely work.

Go back to the equation:

10s = (As + B)(s + 4)(s + 1) + C(s^2 + 4)(s + 1) + D(s^2 +4)(s+4)

Multiply it all out, using the new numbers we found for B, C, and D.

10s = (As + 2)(s + 4)(s + 1) + (2/3)(s^2 + 4)(s + 1) + (-2/3)(s^2 +4)(s+4)

10s = (As + 2)(s^2 + 5s + 4) + (2/3)(s^3 + s^2 + 4s + 4)

+ (-2/3)(s^3 + 4s^2 + 4s + 16)

10s = As^3 + 5As^2 + 4As + 2s^2 + 10s + 8 + (2/3)s^3 + (2/3)s^2

+ (8/3)s + (8/3) -(2/3)s^3 -(8/3)s^2 -(8/3)s – (32/3)

10s = As^3 + (5A + 2 + 2/3 – 8/3)s^2 + (4A + 10)s

So, all we have on the left side is 10s. This means that:

A = 0

(5A + 2 + 2/3 – 8/3) = 0

4A + 10 = 10

The first equation tells us that A = 0. Well, does this hold true in the other equations as well?

5*0 + 2 + 2/3 – 8/3 = 0

2 – 6/3 = 0

2 – 2 = 0. For the second equation, it was true.

4*0 + 10 = 10

10 = 10.

Yes!! Therefore, A = 0, but the conventional methods in addition to our faster method!

So, the answer we get its A = 0, B = 2, C = 2/3, and D = -2/3

If we substitute this in, we get:

10s/((s^2 +4)(s+4)(s+1)) = 2/(s^2 + 4) + 2/(3(s+4)) – 2/(3(s+1))

Which can simplify to:

10s/((s^2 +4)(s+4)(s+1)) = 2/(s^2 + 4) + 2/(3s+12) – 2/(3s+3)

The reason we do all of this is to make our functions integrable. Since we can’t integrate the original function, we try to use partial fractions to separate them in to integrable fractions. However, this one came out interesting. To integrate the first term, you’d have to know that it becomes arctan. I won’t do the integration, because that’s not the problem were working on, but that’s why we do partial fractions in the first place!

So, let me prove to you that this method of partial fractions does work for problems just like this!

I’m just going to change your problem a little bit. Instead of (s^2 + 4), make it (s^2 – 4). This small change allows us to find a zero for this term. Letting s = 2 will make it zero, so now we can move onward!

So, you can find the numbers B, C, and D the exact same way.

You would get B = 50/9, C = 10/9, and D = 10/9.

Now, let s = 2 to find A.

With our change, the equation would be:

10s = (As + B)(s + 4)(s + 1) + C(s^2 – 4)(s + 1) + D(s^2 – 4)(s + 4)

Substitute in 2 for s.

20 = (2A + B)(6)(3)

20 = 36A + 18B

Since B = 50/9,

20 = 36A + 100

-80 = 36A

A = -20/9

See! We can find all of the numbers, even when there is a variable in the numerator.

Any questions on that? I went quickly, rather than explaining every step, because all of the explanation is in the above post!

Thanks for asking question! If you have a question, that means that other people do too! Thank you for the comment!

Mike

## abi says:

October 1, 2011 at 7:07 pm (UTC -5)

very cool!! but it seems to not work for any form having the variable on the numerator? ex. 10s/((s^2 +4)(s+4)(s+1))

## The Rash says:

September 28, 2011 at 6:30 am (UTC -5)

Well, that was so obvious!

But, credits to Mike for making it look so! A genius is someone who makes things look very obvious to others.

Not many would have decoded this simplicity of Partial Fractions, other than memorising the Heaviside Method, which itself originates from this logic!

## Ashish says:

July 22, 2011 at 10:09 am (UTC -5)

Genius man…

Bt the problem is that i have 2 show the older method in my exam sheet…i can use it to quickly check the ans!!thanks….

## Kendra says:

March 5, 2011 at 10:33 pm (UTC -5)

Thank you! This was very helpful.

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