Question

Partial Fractions problem:
Indefinite integral of (2x)/((x+7)^2)

Typically, I would use long division, but it doesn't work here.. or maybe it does, but I am not getting it for some reason.. any help is appreciated!

Answer 1

hmm i dont think partial fractions is needed...just substiution
u = x+7 .... x = u-7
d u = dx

Answer 2

You can avoid having to do anything too messy by just using \(u = x + 7\) and also noting that \(u - 7 = x\), I think.

Answer 3

Well you can use partial fractions

Answer 4

\[\int \frac{2(u - 7)}{u^2}du = \int \frac{2u -14}{u^2}du = \int 2u^{-1} - 14u^{-2} du\]

Answer 5

I figured out how to do it with u substitution, but for my homework - the question asks to show how you can use partial fractions to solve this.

where i got confused was where they separated \[2\int\limits_{}^{}\frac{ 1 }{ x+7 } -\frac{ 7 }{ (x+7)^2 } dx\]

Answer 6

So we would get
\[ \frac{2x}{(x+7)^2}=\frac{A}{x+7}+ \frac{B}{(x+7)^2}\]
\[\frac{2x}{(x+7)^2}=\frac{A(x+7)+B}{(x+7)^2}\]


So \(2x=A(x+7)+B\)

Let x=-7

\[2(-7)=A(-7+7)+B \]
\[-14=B\]

Now let x=0

\[2(0)=A(0+7)-14\]
\[14=7A\]
\[A=2\]

So now we plug our A and B back into our original equation and we get our partial fraction
\[ \frac{2x}{(x+7)^2}=\frac{2}{x+7}-\frac{14}{(x+7)^2}\]

Answer 7

Now we need to differentiate the partial fraction but do you follow how I decomposed the fraction?
Any questions?

Answer 8

Sorry, i'm referencing my book as well.. and I see the rule that states:
A_1/(ax+b) + A_2/(ax+b)^2 +...

i just don't understand the intuition.. if we're separating these fractions, why not decompose the denominator as (x+7)(x+7)?

Answer 9

After doing that, i couldn't solve for A or B, because setting either equal to 0 would cause both to equal 0..

Answer 10

Umm not sure about the intuition gotta think about that one But add the 2 fractions and you will see that you get the result
\[\frac{2x}{(x+7)^2}=\frac{2}{x+7}-\frac{14}{(x+7)^2}\]

Answer 11

haha, ok so I just have to remember that if you have an integral of something over a perfect square, then you do the above?

Answer 12

Well first we wld set up our integral as follows:
\[ \int{\frac{2}{x+7}-\frac{14}{(x+7)^2}dx}\]
\[ \int\frac{2}{x+7}dx- \int\frac{14}{(x+7)^2} dx\]
\[2 \int\frac{1}{x+7}dx- 14\int\frac{1}{(x+7)^2} dx\]

Answer 13

got it.. then substitute for x, with u = x+7

so that's
\[2\int\limits_{}^{}(1/u)du - 14\int\limits_{}^{} (1/u^2)du\]

Answer 14

So far so good :)

Answer 15

2ln(x+7) - 14(x+7)

Answer 16

sorry that's -14(1/u) = -14(1/(x+7))

Answer 17

2ln(x+7) - (14/(x+7))?

Answer 18

wait, thats --, so +14(x+7) .. jeez.. im making so many mistakes

Answer 19

YAAAAAAAAAAAAAAAAA :D

Answer 20

There we go

Answer 21

14/(x+7)*****

Answer 22

ya dw i get it lol

Answer 23

It happens to me all the time plus I am soo slow at latex so it just adds to things

Answer 24

ahhh, this is why i mess up.. i miss the obvious :(

Thank you for your help though, i appreciate the step-by-step process.. it definitely helped!!

Answer 25

But dont forget the +C

Answer 26

oh yeah, when i entered it in online.. they had a 'hint'.. so i didn't forget (otherwise i would have) haha

Answer 27

No problem. Its a refresher for me. I took calc 2 years ago

Answer 28

Great

Answer 29

well if you are looking for more refresher problems i will be here all night :)

Answer 30

lol I may be back. Dinner time though :)

Answer 31

awesome, thanks anyway :D!