anonymous
  • anonymous
Integral of (3x-10)/(x^2+8x+15)? I got (3/2)x^2 -10x + arctan(x+4)+C but it says I'm wrong. Can anyone help me with this?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\int\limits_{}^{} (3x-10)/(x^2+8x+16-1)dx\]\[\int\limits_{}^{} (3x-10)/((x+4)^2-1)dx\]\[\int\limits_{}^{} (3x-10)dx + \int\limits_{}^{} 1/((x+4)^2-1)dx\] This is how I did
TuringTest
  • TuringTest
your mistake is between line 2 and 3
TuringTest
  • TuringTest
\[{3x-10\over(x+4)^2-1}\neq3x-10+{1\over(x+4)^2-1}\]writing what you did clearly like this it should now seem obvious that that was not a legal move ;)

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TuringTest
  • TuringTest
@Dumboy still there?
anonymous
  • anonymous
yeah I see my mistake, I fixed it got to this part \[\int\limits_{}^{} 3x/((x+4)^2-1) dx -\int\limits_{}^{} 10/((x+4)^2-1) dx\] but I'm stuck on integral.. Is this the right way to approach this?
TuringTest
  • TuringTest
sure, there are a number of ways to proceed fro here let's do each integral individually, starting with the second one which can be rewritten\[10\int{dx\over(x+4)^2-1}\]we can use a trig sub for this I think wanna try one?
anonymous
  • anonymous
Yeah that's arctan(x-4) +C but I don't get the first part of the integral. I tried trig substitution but ended up with more completed looking integral :(
TuringTest
  • TuringTest
...that integral above does not come out to arctan(x+4)+C you would need a + in between the terms in the denom for that to work
TuringTest
  • TuringTest
\[\int{dx\over u^2-1}=-\tanh^{-1}u\]notice the \(hyperbolic\) tan there
TuringTest
  • TuringTest
+C
TuringTest
  • TuringTest
yeah these integrals are getting ugly what section are you doing? partial fractions maybe?
anonymous
  • anonymous
yeah partial frations
anonymous
  • anonymous
fractions*
TuringTest
  • TuringTest
ok, then let's back this up and start over, because we are set up to do a trig sub here
TuringTest
  • TuringTest
man those typos are gonna kill me... factor instead of complete the square\[\int{3x-10\over x^2+8x-15}dx=\int{3x-10\over(x+3)(x+5)}dx\]now what do you think? more doable yet?
TuringTest
  • TuringTest
ah I did it again! typo, but I hope you get the point :/
TuringTest
  • TuringTest
\[\int{3x-10\over x^2+8x+15}dx=\int{3x-10\over(x+3)(x+5)}dx\]there!
anonymous
  • anonymous
I guess you could split it into ∫ (3x-10)/(x+3)dx + ∫(3x-10)/(x+5)dx but I'm not sure what to do next
TuringTest
  • TuringTest
no, that's not how partial fractions works at all bro :P first of all that's not even true second of all... brb
anonymous
  • anonymous
is it like this? 3x-10 = A(x-3) + B(x+5)
TuringTest
  • TuringTest
yes, much better but the step before that is important
TuringTest
  • TuringTest
\[{3x-10\over(x+3)(x+5)}=\frac A{x+3}+\frac B{x+5}\]multiply everything by \((x+3)(x+5)\) to arrive at\[3x-10=A(x+5)+B(x+3)\]if you don't do that middle step you will be likely to lose track of which number is \(A\) and which is \(B\)
TuringTest
  • TuringTest
(btw you had x-3 as one of your factors, which is wrong. probably my fault as I think I made that typo earlier)
TuringTest
  • TuringTest
so now that we have\[3x+10=A(x+5)+B(x+3)\]do you know how to determine A and B ?
anonymous
  • anonymous
when x = -3, A = -19/2 and when x = -5, B= 25/2 so (-19/2)(3x-10)/(x+3) +(25/2)(3x-10)/(x+5)?
TuringTest
  • TuringTest
idea is right, math is wrong
anonymous
  • anonymous
oops it's A = 2 and B = -2
TuringTest
  • TuringTest
\[3x+10=A(x+5)+B(x+3)\]\[x=-3:3(-3)+10=A(-3+5)+B(3-3)\]\[1=2A\implies A=\frac12\]
anonymous
  • anonymous
oh..and for B I have -5/2 but shouldn't I get -1/2?
TuringTest
  • TuringTest
yeah I think you should get -1/2 let's see...
TuringTest
  • TuringTest
\[x=-5:3(-5)+10=A(-5+5)+B(-5+3)\]\[-5=-2B\implies B=\frac52\]oh, so we were both wrong :P
anonymous
  • anonymous
\[1/2 \int\limits_{}^{} (3x-10)/(x+3)dx\]\[3x-10 \div x+3 = 3-1/(x+3)\]\[\int\limits\limits_{}^{} 3 - 1/(x+3) dx\]\[3x+\ln(x+3)+C\] is this part correct?
TuringTest
  • TuringTest
no, sorry
anonymous
  • anonymous
I forgot 1/2 in the last part, is that what's wrong or is it other part?
TuringTest
  • TuringTest
I'm sorry but you seem to have no idea what we're doing here you didn't use all the stuff we just did with partial fractions where is the x+5 term? why do you still have 3x-10 ? that should went away when we did the PF part
TuringTest
  • TuringTest
I will demonstrate this one and link you to something I hope that you read thoroughly to clear up these misunderstandings...
TuringTest
  • TuringTest
\[\int{3x-10\over x^2+8x-15}dx=\int{3x-10\over(x+3)(x+5)}dx=\int\frac A{x+3}+\frac B{x+5}dx\]partial fractions:\[A(x+5)+B(x+3)=3x-10\]\[x=-3\implies A=-\frac12\]\[x=-5\implies B=\frac52\]so the integral is now\[-\frac12\int\frac 1{x+3}dx+\frac52\int\frac 1{x+5}dx\]do you understand now?
anonymous
  • anonymous
oh okay so 3x-10 goes away with partial fraction.. I was just doing the first part of the integral up there (forgot minus sign again:P) Thanks

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