Question

calculus 2 - partial fractions

Answer 1

those are fun yet lengthy as the hills

Answer 2

Oh nice, they told you the way to factor it :O that's helpful.
Cubics can be annoying sometimes.

Answer 3

partial fractions happen again for differential equations

Answer 4

Ya that's when they're most important :) Good ole Laplace!

Answer 5

just get the lcd.... and then solve for a b and c

Answer 6

oh wait we get the lcd...I know that... then I think we need to find a b and c and then antiderivative.

Answer 7

So they're telling you that:\[\Large\bf\sf x^3-7x^2+x-7\quad=\quad (x-7)(x^2+1)\]

Answer 8

So for our decomposition setup:\[\Large\bf\sf\frac{114x+2}{x^3-7x^2+x-7}\quad=\quad \frac{A}{x-7}+\frac{Bx+C}{x^2+1}\]Understand why our setup is like that?
We put a polynomial of one degree less in each numerator.

We'll multiply through by the denominator on the left.

Answer 9

\[\Large\bf\sf 114x+2\quad=\quad A(x^2+1)+(Bx+C)(x-7)\]

Answer 10

That step make sense?

Answer 11

yes i think so

Answer 12

We want to solve for A, B and C.
There are some different approaches you can take.
You can multiply everything out and then match up the degrees of x.
You'll be left with a system of equations that's easily solvable.

I prefer another method though.
We just plug in values of x that will allow to quickly solve for one of our unknowns.
In this case, plugging in x=7 will turn that entire second term into 0, right?

Answer 13

\[\Large\bf\sf 114(7)+2\quad=\quad A(7^2+1)+(7B+C)(7-7)\]See how this will allow us to solve for A?

Answer 14

yes

Answer 15

Hmmm so what do you get for A?

Answer 16

123/50?

Answer 17

No you didn't multiply the left side correctly :U

Answer 18

oh iadded

Answer 19

16

Answer 20

ok good good.

Answer 21

We'll plug it in,
\[\Large\bf\sf 114x+2\quad=\quad 16(x^2+1)+(Bx+C)(x-7)\]What value of x will allow us to find B or C?
Hmmmm how bout x=0? See what that will do for us?

Answer 22

bx+C(-7)

Answer 23

2= 16-7c

Answer 24

c=2

Answer 25

Mmm good good!

Answer 26

Again, plug it in,
\[\Large\bf\sf 114x+2\quad=\quad 16(x^2+1)+(Bx+2)(x-7)\]To solve for B, plug in any value for x ( one that we haven't used yet ).
Choose something particularly easy like x=1 or whatever.

Answer 27

116 = 32+(B+2)-6

Answer 28

-16

Answer 29

116 = 32+(B+2)(-6)

^ That's probably what you meant, but use those brackets mister! :O

Answer 30

Ok good we've found our values of A, B and C.

A= 16
B=-16
C= 2

Answer 31

\[\Large\bf\sf \int\limits\frac{114x+2}{x^3-7x^2+x-7}dx\quad=\quad \int\limits\frac{A}{x-7}+\frac{Bx+C}{x^2+1}dx\]

Plugging in our values,\[\Large\bf\sf \int\limits\frac{114x+2}{x^3-7x^2+x-7}dx\quad=\quad \int\limits\frac{16}{x-7}+\frac{-16x+2}{x^2+1}dx\]

Answer 32

From here it should be easier to integrate.

Answer 33

That second term you'll need to split into 2 fractions.\[\Large\bf\sf \int\limits \frac{16}{x-7}+\frac{-16x+2}{x^2+1}dx\quad=\quad \int\limits \frac{16}{x-7}-\frac{16x}{x^2+1}+\frac{2}{x^2+1}dx\]

Answer 34

Confused on how to integrate any of those terms?
Understand how we got to this point? :U

I know the process is a little strange.
Partial Fractions can be tricky.

Answer 35

yea im not good with integration can you explain how to do it there/

Answer 36

\[\Large\bf\sf \int\limits \frac{16}{x}\;dx\]Would you know how to solve this? ^

Answer 37

16lnx?

Answer 38

Yes, good.
Same idea with our problem.

Answer 39

So that first term is giving us 16ln|x-7|

Answer 40

-16xln(x^2+1)+2ln(x^2+1)

Answer 41

I'm confused, what did you do? Parts or something?

Answer 42

The second term should just be a u-substitution.

Answer 43

For the third term,\[\Large\bf\sf \int\limits \frac{2}{x^2+1}\;dx\quad=\quad 2\left(\int\limits \frac{1}{x^2+1}\;dx\right)\quad=\quad 2(?)\]
Remember this integral?
It's a good one to memorize, it shows up a lot.

Answer 44

sry im not good at this

Answer 45

im not sure

Answer 46

8ln(x^2+1)

Answer 47

ok good.

Answer 48

what now

Answer 49

@zepdrix

Answer 50

Im not sure what the answer is. The options look different than what I got

Answer 51

16lnx+ln(x^2+1)+2arctanx

Answer 52

Hmm, where did the 8 go on the middle term?
And the negative sign?

Answer 53

And we had 16ln|x-7| didn't we? How did that change to 16lnx? :(

Answer 54

oh
16ln|x-7|-8ln(x^2+1)+2arctanx

Answer 55

Wow they really went out of their way to make the answer choices very difficult to get to.... annoying :(

Answer 56

So on the middle term, we have to do something fancy.

Answer 57

\[\Large\bf\sf -8\ln|x^2+1|\quad=\quad -16\cdot\frac{1}{2}\ln|x^2+1|\]Then bring the 1/2 inside using rules of logs.\[\Large\bf\sf -16\ln\sqrt{x^2+1}\]1/2 power is the square root.

Answer 58

so the third option?

Answer 59

Then use your rules of logs to combine the first two logs.\[\Large\bf\sf \log(a)-\log(b)\quad=\quad \log\left(\frac{a}{b}\right)\]

Answer 60

Mmm ya it looks like it's probably the third option, lemme check wolfram a sec

Answer 61

Mmm ya looks like the third option!
Sheesh, tough problem!

Answer 62

thanks a lot