anonymous
  • anonymous
Need help integrating (x+1)/(x^3 + x).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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slaaibak
  • slaaibak
\[{x+1} \over {x(x^2 + 1)}\] Use partial fractions.
anonymous
  • anonymous
I tried using partial fractions, and got A(x^2 + 1) + (Bx + C)x, but I'm not sure how to get the values of the coefficients.
TuringTest
  • TuringTest
\[A(x^2+1)+(Bx+C)x=x+1\to (A+B)x^2+Cx+A\]just looking at the coefficients it looks like A=1, C=1, and A+B=0 wich leads to B=-1

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slaaibak
  • slaaibak
\[x + 1 = A(x^2 + 1) + (Bx + C) x\] \[x + 1 = (A+B)x^2 + Cx + A\] Therefore A + B = 0 C=1 A=1 damn turing beat me :/
TuringTest
  • TuringTest
sorry, you got your medal though ;)
slaaibak
  • slaaibak
We will continue this battle in another question ;)
TuringTest
  • TuringTest
indeed! did you get the idea Xylienda?
anonymous
  • anonymous
:) Thanks guys!
anonymous
  • anonymous
Yeah. So its more like a trial and error thing, right?
TuringTest
  • TuringTest
Not really in this case when you spell it out and look at the coefficient on the right side you get\[(A+B)x^2+Cx+A=x+1\]the coefficient of x^2 on the right is zero, so\[A+B=0\]the coefficient on x is 1, so\[C=1\]and the constant term is just 1, so\[A=1\]now that we know A=1 we can go back and solve the first one\[A+B=0\to B=-1\]so no, it's not trial and error. It is making a system from by setting the coefficients on each side equal. See?
anonymous
  • anonymous
Oh! I think i really get it now. So you kind of begin by, more or less, matching up the forms on both sides of the equation and compare the coefficients, then work from there.
TuringTest
  • TuringTest
exactly collect like terms on the left and set each coefficient equal to that on the right. Sometimes the systems will be more complex, but here it is pretty easy.
anonymous
  • anonymous
Thank you! :)

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