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unimatix
 one year ago
Calculate the integral
Integral x/((xa)(xb)) dx for
a) a /= b (a does not equal b)
b) a = b
unimatix
 one year ago
Calculate the integral Integral x/((xa)(xb)) dx for a) a /= b (a does not equal b) b) a = b

This Question is Closed

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.2Partial fraction decomposition?

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0I know. I tried. But when I go to try and find anything I can't get anything. For example part a.dw:1441687142216:dw

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441687204033:dw

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441687237178:dw

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0but that seems pretty useless

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{align} \frac1{(xa)(xb)}&=\frac A{xa}+\frac B{xb}\\ 1&=A(xb)+B(xa) \end{align}\] For \(x=a\) \[1=A(ab)\implies A=\frac1{ab}\] For \(x=b\) \[1=B(ba)\implies B=\frac1{ba}\] Therefore \[\begin{align} \frac1{(xa)(xb)}&=\frac 1{(ab)(xa)}+\frac 1{(ba)(xb)}\\[2ex] \frac x{(xa)(xb)}&=\frac x{(ab)(xa)}\frac x{(ab)(xb)}\\ &=\frac1{ab}\left(\frac x{xa}\frac x{xb}\right)\\ \end{align}\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.2@UnkleRhaukus why is it \(\dfrac{1}{(xa)(xb)}\) instead of \(\dfrac{x}{(xa)(xb)}\)?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0You mean when he starts the decomp?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.2Just to make it easier to evaluate?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Well notice that right after he says Therefor, he states the thing with 1 in the numerator, but then multiplies that result by x to get the desired equality

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0yes, to make it easier :)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.2Would it have changed the outcome in solving for \(A\) and \(B\) if he were to include \(x\) in the decomp process ?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0there would have been an x next to some things.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0exactly every \(1\) would be an \(x\), except the last one in the post :)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.2That didnt make sense :\

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0look at U.R. post, everywhere you see a \(1\)(except the last one) , replace is with an \(x\).

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.2Oh, yeah that makes sense.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{align} \frac x{(xa)(xb)}&=\frac{Ax}{xa}+\frac{Bx}{xb}\\ 1&=Ax(xb)+Bx(xa) \end{align}\] For \(x=a\)\[1=Aa(ab)\implies A=\frac1{a(ab)}\] For \(x=b\)\[1=Bb(ba)\implies B=\frac1{b(ba)}\] Therefore: \[\begin{align} \frac x{(xa)(xb)} &= \frac x{a(ab)(xa)}+\frac x{b(ba)(xb)}\\ &= \frac1{ab}\left(\frac x{xa}\frac x{xb}\right) \end{align}\] @Jhannybean This is the same result; decomposing with the numerator term. In my opinion it is much harder to follow.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.2Ok, yeah it does seem more complicated.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1@unimatix, can you integrate now?

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0Yep thanks for the help!

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.2@unimatix have you looked more into that other problem you were working on? Figured it out?
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