## sjg13e one year ago Another Partial Functions Problem

1. sjg13e

2. sjg13e

$\int\limits_{}^{}\frac{ 4 }{ x^2 + 5x - 14 }dx$

3. anonymous

$\frac{A}{x+7}+\frac{B}{x-2}$ is a start

4. sjg13e

$\int\limits_{}^{}\frac{ 4 }{ x^2 + 5x - 14 }dx = \int\limits_{}^{}\frac{ 4 }{ (x-1)(x+1) }dx = \frac{ A }{ x+7 } + \frac{ B }{ x-2 }$

5. anonymous

you want to find A and B the snappy way?

6. anonymous

or the slower way (same thing really)

7. sjg13e

I was told to find A and B by doing system of equations. I do know that you can also find A and B quickly by letting x = -7 and x = 2, then solving for A and B. But let's do it the first way because that's where I messed up

8. anonymous

lets do it the real snappy way first

9. sjg13e

Okay

10. sjg13e

The way I did it: 4 = A(x-2) + B(x+7) 4 = Ax - 2A + Bx + 7B 4 = (A+B)x - 2A + 7B 4 = -2A - 7B A + B = 0 4 = -2(-B) - 7B A = -B 4 = 2B - 7B A = -(-4/5) 4 = -5B A = 4/5 -4/5 = B

11. anonymous

$\frac{4}{(x+7)(x-2)}=\frac{A}{x+7}+\frac{B}{x-2}$ to find $$A$$ put your finger over the factor of $$x+7$$ in $\frac{4}{(x+7)(x-2)}$and replace $$x$$ by $$-7$$ $\frac{4}{\cancel{(x+7)}(-7-2)}$

12. anonymous

you get $A=-\frac{4}{9}$ instantly

13. anonymous

you want to do it your "system of equations" long boring way and see if we can find the mistake?

14. sjg13e

Then, $\int\limits_{}^{}\frac{ 4 }{ x^2 + 5x - 14 }dx = \int\limits_{}^{}\frac{ 4 }{ 5(x+7) }dx - \int\limits_{}^{}\frac{ 4 }{5(x-2) }dx$ So the answer I got is $= \frac{ 4 }{ 5 }\ln |x+7| - \frac{ 4 }{ 5 }\ln |x-2| + C$

15. anonymous

yeah the 5 is wrong, should be 9

16. sjg13e

I understand the method you used! Yes, can we try the boring way? Lol. I just want to make sure I can do both methods

17. sjg13e

Yeah, I don't know where I went wrong. Somehow I ended up with 5 instead of 9

18. anonymous

ok i like the "put your finger over the factor" method a lot more

19. sjg13e

Haha, me too! But I need to be familiar with both methods for my exam

20. anonymous

ok sure lets see the mistake $A(x-2)+B(x+7)=4$is a start then \$Ax-2A+Bx+7B=4$so $A+B=0\\-2A+7B=4$

21. sjg13e

Oh wait lol i see it

22. anonymous

oh i see it right away since $$A=-B$$ you have $2B+7B=4\\ B=\frac{4}{9}$

23. sjg13e

I wrote 4 = -2A - 7B instead of 4 = -2A + 7B.

24. anonymous

ooh ok well glad we cleared that up!

25. sjg13e

then A = -4/9 and B = 4/9 and then we get the right answer haha. Thank you!

26. anonymous

yw

27. anonymous

there was a fellow here a while ago "eliasaab" i think, who had an amazingly snappy way to do this, even if you had annoying repeated roots i forget what it was though, had to do with derivatives

28. sjg13e

nice, maybe i'll come across the method again

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