how to solve partial fractions

This math tutorial will help you understand how to solve for unknown constants when doing partial fractions. This will help you with your precalculus and calculus 2 classes.

0:00 1st example
4:19 2nd example
8:42 3rd example
11:14 4th example
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[alaeGhemra]

I don't know how I can thank you. You really deserve to be a professor

[fantiscious]

It's wild that cover-up method is not taught in schools, it's one of the most awesome shortcuts ever. At least better than those "shortcuts" tiktok keeps showing me to multiply two digit numbers lol

[vaishnavimajumdar]

he is the best maths teachers out there, like seriously maths can never be this easy but his teaching makes it sound so simple.

[lucasrxtx]

Guy does a little stretch at the beginning, then I see a whole shelf of expo markers in the background... Oh man, here we go, this guy is about to wreck my mind 🤯😅 Awesome video, thanks for the content 🙏

[MohamedMoncifFOKROUN-vp]

you are contributing to make the world a better place by spreading such knowldge to everyone for free

[carultch]

Other tips for examples 2 thru 4:

For example 2, plug in strategic values of x, such as x=0 and x=1, that aren't already spoken-for by the coverup method.
(2*x + 1)/((x + 1)*(x^2 + 2)) = -1/3/(x + 1) + (B*x + C)/(x^2 + 2)

Let x = 0, to eliminate B and find C:
(1)/((1)*(2)) = -1/3/(1) + (C)/(2)
1/2 + 1/3 = C/2
C = 5/3

(2*1 + 1)/((1 + 1)*(1 + 2)) = -1/3/(1 + 1) + (B + 5/3)/(1 + 2)
1/2 = -1/6 + B/3 + 5/9
B = 1/3

For examples 3 & 4:
Once Cover-up finds A and C, you can find B, by multiplying by only 1 copy of the repeated term. Take the limit as x goes to infinity. Some terms cancel to zero, some terms reduce to a finite non-zero number, and B will stand on its own as a constant. This is a way to more directly find B, that allows you to ignore terms that don't affect it. It has trouble if there is an irreducible quadratic involved, but works well when only dealing with linear and linear-repeated factors.

As applied to example 3:
(2*x + 1)/((x + 1)*x^2) = -1/(x + 1) + B/x + 1/x^2

Multiply by just one copy of x:
(2*x + 1)/((x + 1)*x) = -x/(x + 1) + B + 1/x

Send x to inf, and take the limit:
0 = -1 + B + 0
B = 1

[GoChelsea-vquq]

You know a teacher is GOATED when he uses two markers with one hand 🐐

[bigsmoke-theoneandonly-man]

Was needing to catch up on some specialist mathematics (Calculus 2 I guess) and I had no idea how to rationalise the 3rd/4th example using that form of partial fraction. This demonstration was so clear and now I have a simple way to rationalise and complete these problems, thanks so much!

[Calderay]

you really helped me so much. Thank you. Im not an english native speaker but i understood everything so well. Congratulations ❤️

[rssl]

Thank you so much! I’m preparing for a competition and I knew something I this existed for these type of questions but I actually forgot the method <3

[DarkNight]

Thank you very much this video really helped

[mohamadothman]

you are amazing at explaning, habibi <3

[Silver-cuup]

Dear Blackpenredpen,
Can you explain why lim x->inf [xtanh(x)-ln(cosh(x))] = ln(2)?
There is something exponentially weird with this limit.

[GottfriedLeibniz]

1st example is also called Heaviside method

[RujitStha]

do the one where the upper term also has x squared

[abhijoyghosh]

I wish you could give us a pdf sheet related to the concept every video 😢

[AkinmusereTimileyin]

Thanks for this but can you help me with this x/(x^2+a^2)(x^2+b^2).😊😊

[DeLuini]

Thank you so much from Croatia. Great help and excellent explanation of why we can do this.
May our God Jesus Christ of Nazareth bless you and keep you.

[jordimayorgisbert]

I'm thinking is the fastest method. Am I wrong? Thank you for your awesome work, @bprp !!!

[abanobkute]

10q but No 94, when we solve by shortway like for A we did x=-1 and found A=-1 and why you didn't use for B, (x=-2) and like this # u did that for C and by long way for B, then why ?