anonymous
  • anonymous
int. dx/ (x^2 +2x+10) using partial fraction. please.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
.Sam.
  • .Sam.
you can't factor that
.Sam.
  • .Sam.
best approach is to complete the square and then trig substitution
wasiqss
  • wasiqss
yehh, it will have complex factors

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anonymous
  • anonymous
yup
anonymous
  • anonymous
@lilMissMindset To complete the square x^2 +2x+10= x^2+2x+1+9 = (x+1)^2 +3^2 \[\int\limits_{}^{}\frac{dx}{(x+1)^2+3^2}\] Now complete the rest
anonymous
  • anonymous
Remember \[\int\limits_{}^{}\frac{dt}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a})\]
anonymous
  • anonymous
You can eaily prove the above by substituting x = atan(theta)
anonymous
  • anonymous
@shivam_bhalla she said partial fraction
TuringTest
  • TuringTest
but it cannot be done with partial fraction if it cannot be factored
wasiqss
  • wasiqss
traile , it willl become very complex then
anonymous
  • anonymous
yeah, im supposed to use that, partial fraction
anonymous
  • anonymous
@LOL, whjy do you want to complicate things when there is a aeasier method available. If you still insist on partial fraction, then it is fine
wasiqss
  • wasiqss
lil miss it will only have complex factors
TuringTest
  • TuringTest
there is either a typo, or it's impossible
anonymous
  • anonymous
*why
TuringTest
  • TuringTest
I mean that using partial fractions is impossible... or at least redundant
anonymous
  • anonymous
what am i going to do about this problem then?
anonymous
  • anonymous
Take x+1 = t dx=dt You still get the same thing with partial fractions too
TuringTest
  • TuringTest
how would you do this with partial fractions? I don't see it... perhaps I am wrong though, it wouldn't be the first time :P
anonymous
  • anonymous
do you think quadratic factors can be use
anonymous
  • anonymous
\[\frac{1}{t^2+9} = \frac{At+B}{t^2+9}\] we see a = 0, B=1 We again get back the same thing. So partail fraction approach should be useless
anonymous
  • anonymous
*partial
TuringTest
  • TuringTest
like I said then, it's just redundant
TuringTest
  • TuringTest
by impossible, I meant that the operation is useless, as you just said
wasiqss
  • wasiqss
you fail turing xD lol jk
TuringTest
  • TuringTest
@lilMissMindset are you \(sure\) there isn't a typo in your post?
anonymous
  • anonymous
yea. im quite sure i typed it right.
anonymous
  • anonymous
let's ignore the partial fraction thingy then.
TuringTest
  • TuringTest
then shivam bhalla has shown you the right way do you know trig substitution integrals?
anonymous
  • anonymous
yeah. i'm sorry, i'll do it using trig substitution. thank you so much.
TuringTest
  • TuringTest
welcome!

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