Question

Partial Integration

Answer 1

\[\int\limits \frac{ 4x ^{2} }{ x ^{3}+x ^{2}-x-1 } dx\]

Answer 2

factor out the bottom by (x-1)(x+1)^2?

Answer 3

Mmm ok sounds good so far.
Gotta break it down using Partial Fraction Decomposition from there?

Answer 4

\[\frac{ A }{ (X-1) } +\frac{ BxC }{ (x+1)^{2} } ?\]

Answer 5

lower case x i meant for denominator of A

Answer 6

No no.
You have a repeated linear factor.
Think of (x+1)^2 as (x+1)(x+1),
\[\Large\rm \frac{4x^2}{(x-1)(x+1)^2}=\frac{A}{x-1}+\frac{B}{(x+1)}+\frac{C}{(x+1)^2}\]

Answer 7

Here is another example:\[\Large\rm \frac{stuff}{(x-2)^4}=\frac{A}{(x-2)}+\frac{B}{(x-2)^2}+\frac{C}{(x-2)^3}+\frac{D}{(x-2)^4}\]

Answer 8

You have to have a term for every power leading up to the .. power.
I can't exactly remember .. why though >.<

Answer 9

okay got it so no I have to take the denominators that each don't have so

then im left with
4x^2 = A(X+1) +B(x-1) + C(x+1)^2 then distribute?

Answer 10

or for C is it (x+1) also

Answer 11

*now

Answer 12

Hmm that doesn't look right...
You're multiplying through by the denominator on the left,\[\rm \color{royalblue}{\left(\frac{(x-1)(x+1)^2}{(x-1)(x+1)^2}\right)}\frac{4x^2}{(x-1)(x+1)^2}=\left[\frac{A}{x-1}+\frac{B}{(x+1)}+\frac{C}{(x+1)^2}\right]\color{royalblue}{\left(\frac{(x-1)(x+1)^2}{(x-1)(x+1)^2}\right)}\]You have a bunch of cancellations in each term,\[\Large\rm 4x^2=A(x+1)^2+B(x-1)(x+1)+C(x-1)\]

Answer 13

Ahh woops, you're multiplying each side by (x-1)(x+1)^2, i didn't mean to make that a fraction :( shouldn't be a bottom in the blue thing.

Answer 14

\[\rm \color{royalblue}{(x-1)(x+1)^2}\frac{4x^2}{(x-1)(x+1)^2}=\left[\frac{A}{x-1}+\frac{B}{(x+1)}+\frac{C}{(x+1)^2}\right]\color{royalblue}{(x-1)(x+1)^2}\]This is what I meant to write.

Answer 15

So if you look at the C term,
after distributing you get \(\Large\rm \dfrac{C(x-1)(x+1)^2}{(x+1)^2}\)
Understand how I simplified things down?

Answer 16

@zepdrix using \(\dfrac{Bx+C}{(x+1)^2}\) is fine, you end up getting \(B=0\). This just happens to be the case with all repeated factors.

Answer 17

so is A term written A(x+1)/(x-1) + B(x-1)(x+1) + C(x-1)/(x+1)^2 ?

Answer 18

\[\frac{ A(x+1) }{ (x-1) (x+1) } + \frac{ B(x-1) }{ (x+1)(x-1) } + \frac{ C(x-1) }{ (x+1)^{2}(x-1) }\]

Answer 19

then the denominators would cancel out and i can distribute from there to find my A,B,and C?

Answer 20

@zepdrix

Answer 21

I don't understand what you just wrote :(

Answer 22

so how is it done exactly when you distribute so you get the A,B,and C?

Answer 23

You're multiplying like this:
\[\left[\frac{A}{x-1}+\frac{B}{(x+1)}+\frac{C}{(x+1)^2}\right]\color{royalblue}{(x-1)(x+1)^2}\]to get rid of the denominator on the left side of the equation.
So you get:\[\frac{A\color{royalblue}{(x-1)(x+1)^2}}{x-1}+\frac{B\color{royalblue}{(x-1)(x+1)^2}}{(x+1)}+\frac{C\color{royalblue}{(x-1)(x+1)^2}}{(x+1)^2}\]From there you get cancellations:\[\frac{A\cancel{(x-1)}(x+1)^2}{\cancel{x-1}}+\frac{B(x-1)(x+1)^{\cancel21}}{\cancel{(x+1)}}+\frac{C(x-1)\cancel{(x+1)^2}}{\cancel{(x+1)^2}}\]

Answer 24

ooo OKAY

\[4x ^{2} = A(x+1)^2+B(x-1)+C(x-1)\]

Answer 25

so 4x^2 = Ax^2+A+Bx-B+Cx-C ?

Answer 26

No the middle term still has a factor of (x+1), see it?
There was a square in the numerator, so we lost one of them. Not both.

Answer 27

And also, DON'T EXPAND EVERYTHING OUT.
There is an easier way to finish it up from here.

Answer 28

okay i see it and oo alright my professor usually expands it out

Answer 29

\[\Large\rm 4x^{2} = A\color{orangered}{(x+1)}^2+B(x-1)\color{orangered}{(x+1)}+C(x-1)\]See how the A and B terms both have this factor of x+1?
If we could make the A and B term disappear, we could solve for C easily.

Answer 30

oo ok

Answer 31

ok so can u just show me how its done so i can see the way you solve because my professor explains another way similar but different

Answer 32

We're going to pick specific x values to zero them out.
So when x=-1 we get,\[\Large\rm 4(-1)^{2} = A\color{orangered}{(-1+1)}^2+B(-1-1)\color{orangered}{(-1+1)}+C(-1-1)\]See how the A and B terms will go away because of the orange parts?

Answer 33

Leading to,\[\Large\rm 4=0+0-2C\]

Answer 34

I saw that I had a factor of (x+1), so I specifically chose x to equal -1 so that factor would become zero. I hope it makes sense why I chose that value.

Answer 35

Then solving for C is pretty straight forward,\[\Large\rm 4=-2C\]\[\Large\rm C=-2\]

Answer 36

We'll plug that value back in,\[\Large\rm 4x^{2} = A(x+1)^2+B(x-1)(x+1)-2(x-1)\]To solve for A, notice that the second and third terms both have a factor of (x-1).
So we'll plug in x=1.

Answer 37

\[\Large\rm 4(1)^{2} = A(1+1)^2+0+0\]Which will zero out our B and C because of the (1-1) factor.

Answer 38

Leading to,\[\Large\rm A=1\]

Answer 39

\[\Large\rm 4x^{2} = (x+1)^2+B(x-1)(x+1)-2(x-1)\]To solve for B, just choose something easy to work with.
Any x value that you haven't used up to this point will work.
Plugging in x=0 gives us,\[\Large\rm 4(0)^{2} = (1)^2+B(-1)(1)-2(-1)\]

Answer 40

Leading to,\[\Large\rm 0=0-B+2\]\[\Large\rm B=2\]

Answer 41

I guess I shouldn't say it's a ton easier, I find it easier though.
If you still prefer expanding everything out though, that's fine.

Answer 42

it is a lot easier that way i would lessen my chances of making a mistake

Answer 43

\[\Large\rm \frac{4x^2}{(x-1)(x+1)^2}=\frac{A}{x-1}+\frac{B}{(x+1)}+\frac{C}{(x+1)^2}\]Wolfram is actually telling me that the factors are supposed to be:
A=1
B=3
C=-2

So I made a mistake on the B term.

Answer 44

Ah yes, I see.\[\Large\rm 4(0)^{2} = (1)^2+B(-1)(1)-2(-1)\]\[\Large\rm 0 = 1-B+2\]\[\Large\rm B=3\]Ok ok fixed :3

Answer 45

\[\int\limits \frac{ 1 }{ (x-1) }+\int\limits \frac{ 3 }{ (x+1)}+\int\limits \frac{ -2 }{ (x+1)^{2} }\]

Answer 46

so ln(x-1)+3(x+1) -2ln(x+1)^2

Answer 47

Mmmmm no....

Answer 48

The first two are giving you logs.
I'm not sure what happened with your second integral.. it should give you a log, yes?

Answer 49

i meant 3ln(x+1)

Answer 50

But the third one, no log.
Recall:\[\Large\rm \int\limits \frac{1}{u^2}du=\int\limits u^{-2}du=\frac{1}{-2+1}u^{-2+1}=-\frac{1}{u}\]

Answer 51

So:\[\Large\rm =\ln(x-1)+3\ln(x+1)+~?\]

Answer 52

2/x+1+c

Answer 53

We just simply apply our power rule to the last term.\[\int\limits\limits \frac{ 1 }{ (x-1) }+\int\limits\limits \frac{ 3 }{ (x+1)}+\int\limits\limits \frac{ -2 }{ (x+1)^{2} }\]\[\ln(x-1)+3\ln(x+1)+\frac{-1(-2)}{x+1}\]Mmm ya good!

Answer 54

thank you so much for your patience an help! helped a lot

Answer 55

Np!