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

1. Jhannybean

Partial fraction decomposition?

2. unimatix

I know. I tried. But when I go to try and find anything I can't get anything. For example part a.|dw:1441687142216:dw|

3. unimatix

|dw:1441687204033:dw|

4. unimatix

|dw:1441687237178:dw|

5. unimatix

but that seems pretty useless

6. UnkleRhaukus

\begin{align} \frac1{(x-a)(x-b)}&=\frac A{x-a}+\frac B{x-b}\\ 1&=A(x-b)+B(x-a) \end{align} For $$x=a$$ $1=A(a-b)\implies A=\frac1{a-b}$ For $$x=b$$ $1=B(b-a)\implies B=\frac1{b-a}$ Therefore \begin{align} \frac1{(x-a)(x-b)}&=\frac 1{(a-b)(x-a)}+\frac 1{(b-a)(x-b)}\\[2ex] \frac x{(x-a)(x-b)}&=\frac x{(a-b)(x-a)}-\frac x{(a-b)(x-b)}\\ &=\frac1{a-b}\left(\frac x{x-a}-\frac x{x-b}\right)\\ \end{align}

7. Jhannybean

@UnkleRhaukus why is it $$\dfrac{1}{(x-a)(x-b)}$$ instead of $$\dfrac{x}{(x-a)(x-b)}$$?

8. zzr0ck3r

You mean when he starts the decomp?

9. Jhannybean

Yeah.

10. Jhannybean

Just to make it easier to evaluate?

11. zzr0ck3r

Well 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

12. zzr0ck3r

yes, to make it easier :)

13. Jhannybean

Would it have changed the outcome in solving for $$A$$ and $$B$$ if he were to include $$x$$ in the decomp process ?

14. zzr0ck3r

there would have been an x next to some things.

15. zzr0ck3r

exactly every $$1$$ would be an $$x$$, except the last one in the post :)

16. Jhannybean

That didnt make sense :\

17. zzr0ck3r

look at U.R. post, everywhere you see a $$1$$(except the last one) , replace is with an $$x$$.

18. Jhannybean

Oh, yeah that makes sense.

19. UnkleRhaukus

\begin{align} \frac x{(x-a)(x-b)}&=\frac{Ax}{x-a}+\frac{Bx}{x-b}\\ 1&=Ax(x-b)+Bx(x-a) \end{align} For $$x=a$$$1=Aa(a-b)\implies A=\frac1{a(a-b)}$ For $$x=b$$$1=Bb(b-a)\implies B=\frac1{b(b-a)}$ Therefore: \begin{align} \frac x{(x-a)(x-b)} &= \frac x{a(a-b)(x-a)}+\frac x{b(b-a)(x-b)}\\ &= \frac1{a-b}\left(\frac x{x-a}-\frac x{x-b}\right) \end{align} @Jhannybean This is the same result; decomposing with the numerator term. In my opinion it is much harder to follow.

20. Jhannybean

Ok, yeah it does seem more complicated.

21. UnkleRhaukus

@unimatix, can you integrate now?

22. unimatix

Yep thanks for the help!

23. Jhannybean

@unimatix have you looked more into that other problem you were working on? Figured it out?