sjg13e
  • sjg13e
Another Partial Functions Problem
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
sjg13e
  • sjg13e
please give me second to upload the problem and my work
sjg13e
  • sjg13e
\[\int\limits_{}^{}\frac{ 4 }{ x^2 + 5x - 14 }dx\]
anonymous
  • anonymous
\[\frac{A}{x+7}+\frac{B}{x-2}\] is a start

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sjg13e
  • sjg13e
\[\int\limits_{}^{}\frac{ 4 }{ x^2 + 5x - 14 }dx = \int\limits_{}^{}\frac{ 4 }{ (x-1)(x+1) }dx = \frac{ A }{ x+7 } + \frac{ B }{ x-2 }\]
anonymous
  • anonymous
you want to find A and B the snappy way?
anonymous
  • anonymous
or the slower way (same thing really)
sjg13e
  • 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
anonymous
  • anonymous
lets do it the real snappy way first
sjg13e
  • sjg13e
Okay
sjg13e
  • 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
anonymous
  • 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 \(-7\) \[\frac{4}{\cancel{(x+7)}(-7-2)}\]
anonymous
  • anonymous
you get \[A=-\frac{4}{9}\] instantly
anonymous
  • anonymous
you want to do it your "system of equations" long boring way and see if we can find the mistake?
sjg13e
  • 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 \[= \frac{ 4 }{ 5 }\ln |x+7| - \frac{ 4 }{ 5 }\ln |x-2| + C\]
anonymous
  • anonymous
yeah the 5 is wrong, should be 9
sjg13e
  • 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
sjg13e
  • sjg13e
Yeah, I don't know where I went wrong. Somehow I ended up with 5 instead of 9
anonymous
  • anonymous
ok i like the "put your finger over the factor" method a lot more
sjg13e
  • sjg13e
Haha, me too! But I need to be familiar with both methods for my exam
anonymous
  • anonymous
ok sure lets see the mistake \[A(x-2)+B(x+7)=4\]is a start then \\[Ax-2A+Bx+7B=4\]so \[A+B=0\\-2A+7B=4\]
sjg13e
  • sjg13e
Oh wait lol i see it
anonymous
  • anonymous
oh i see it right away since \(A=-B\) you have \[2B+7B=4\\ B=\frac{4}{9}\]
sjg13e
  • sjg13e
I wrote 4 = -2A - 7B instead of 4 = -2A + 7B.
anonymous
  • anonymous
ooh ok well glad we cleared that up!
sjg13e
  • sjg13e
then A = -4/9 and B = 4/9 and then we get the right answer haha. Thank you!
anonymous
  • anonymous
yw
anonymous
  • 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
sjg13e
  • sjg13e
nice, maybe i'll come across the method again

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