anonymous
  • anonymous
Evaluate the integral: Indefinite Integral ((x^3 + x^2 + 2x + 1) / (x^2 + 1)(x^2 + 2)) dx
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
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anonymous
  • anonymous
so far I have x^3(A+C) + x^2 (B+D) + x(2A +C) +(2B + D)
anonymous
  • anonymous
But I believe I'm going in the wrong direction

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slaaibak
  • slaaibak
\[{{x^3 + x^2 + 2x + 1}\over{(x^2 + 1)(x^2 + 2)} } = {{Ax + B}\over{x^2 + 1}} + {{Cx + D}\over{x^2 + 2}}\] By partial franctions. Now:
anonymous
  • anonymous
ok
anonymous
  • anonymous
I'm good so far
slaaibak
  • slaaibak
\[x^3 + x^2 + 2x + 1 = (Ax + B)(x^2 + 2) + (Cx + D)(x^2 + 1)\] \[x^3 + x^2 + 2x + 1 = Ax^3 + 2Ax + Bx^2 + 2B + Cx^3 + Cx + Dx^2 + D\] \[x^3 + x^2 + 2x + 1 = (A + C)x^3 + (B + D)x^2 + (2A + C)x + (2B + D)\] A + C = 1 B + D = 1 2A + C = 2 2B + D = 1
anonymous
  • anonymous
Alright...but the answer I saw was (1/2)ln (x^2 + 1) + (1 / square(2)) arctan(x/square(2)) + C
anonymous
  • anonymous
It's a frustrating problem
slaaibak
  • slaaibak
Lets take it further then: By solving A, B, C, D: A = 1 C=0 B=0 D=1
anonymous
  • anonymous
ok
slaaibak
  • slaaibak
\[{{x^3 + x^2 + 2x + 1}\over{(x^2 + 1)(x^2 + 2)}} = {x \over {x^2 + 1} } + {{1}\over{x^2 + 2}}\]
slaaibak
  • slaaibak
Now the fun starts! Now we integrate each one of those
anonymous
  • anonymous
ohhhhh ok
slaaibak
  • slaaibak
Now to integrate the first one: \[\int\limits {x \over {x^2 + 1} } dx\] \[x^2 = u\] \[2x dx = du\] \[xdx ={1\over2}du\] \[{1\over2}\int\limits{1\over{u + 1}} du\] =\[{1\over2}\ln(u + 1) + C\] resubstituting u with x^2: \[{1\over2}\ln(x^2 + 1) + C\]
slaaibak
  • slaaibak
Do you need help with the second part of the integral?
anonymous
  • anonymous
please if you don't mind
slaaibak
  • slaaibak
\[\int\limits{1\over{x^2 + 2}} dx\] now, by factoring out the 2: \[\int\limits {1\over{2({1\over2}x^2 + 1)}} dx\] \[{1\over2}\int\limits{1\over{({1\over2}x^2 + 1)}}dx\] Putting the 1/2 inside the bracket: \[{1\over2} \int\limits { {1}\over{ ({{x}\over{\sqrt2}})^2 + 1 } }dx\] \[{x \over {\sqrt 2} } = u\] \[dx = \sqrt2 du\] \[{1\over2}\int\limits {{\sqrt2}\over{u^2+1}} du\] \[{{\sqrt2}\over2}\int\limits {{du}\over{u^2+1}}\] = \[{\sqrt2 \over 2} \arctan u + c\] substituting back: \[{\sqrt2 \over 2} \arctan({x \over \sqrt2}) + C\] since \[{\sqrt2 \over 2 } = {1\over \sqrt2}\] \[{1\over \sqrt2} \arctan({x \over \sqrt2}) + C\]

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