anonymous
  • anonymous
complete the following integral or partial substitution with a. ∫(x^5+2x)/(x^6+6x^2+56)^2 dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
∫(x^5+2x)/(x^6+6x^2+56)^2 dx ∫(x^5+2x) dx - ∫(x^6+6x^2+56)^-2 dx
anonymous
  • anonymous
You can't do this in products, only in additions. suzi20
dumbcow
  • dumbcow
u = x^6 +6x^2 +56 du = 6x^5 +12x dx du = 6(x^5 +2x) dx

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anonymous
  • anonymous
dv = (x^6+6x^2+56)^-2 dx v = -1 (x^6+6x^2+56)^-1 uv - ∫ v du (x^6 +6x^2 +56)(-(x^6+6x^2+56)^-1) - 1/6∫(-(x^6+6x^2+56)^-1) (x^5 +2x) dx hmm... hard
anonymous
  • anonymous
Suzi please continue the answer
anonymous
  • anonymous
u have to do partial again
anonymous
  • anonymous
Let x^6 + 6x^2 + 56 = t Now, du = 6x^5 + 12x dx = 6(x^5 + 2x) dx The integral becomes, \[1/6\int\limits_{}{}dt/t^2\] \[-1/6t\] Putting back the value of t = -1/6(x^6+6x^2+56)
anonymous
  • anonymous
I could ask you to complete
anonymous
  • anonymous
that's calculus 2, i remember now, wow amogh cool...........
anonymous
  • anonymous
amogh can I ask you to write down the answer from the first until the end if you're willing I will be very thankful to you
anonymous
  • anonymous
he did
anonymous
  • anonymous
Tell me what you didn't understand, I've written it completely!
anonymous
  • anonymous
oh I'm sorry, I really do not understand about the calculus lesson, but I want to learn
anonymous
  • anonymous
Let x^6 + 6x^2 + 56 = t Differentiating on both sides, Now, dt = (6x^5 + 12x)dx = 6(x^5 + 2x) dx Multiplying the integral by 6 and diving by 6, \[1/6\int\limits_{}^{}6(x^5+2x)/(x^6+6x^2+56)^2 dx\] Now 6(x^5+2x) becomes dt. The integral becomes, \[1/6\int\limits_{}^{} dt/t^2\] = −1/6t Putting back the value of t = -1/6(x^6+6x^2+56)
anonymous
  • anonymous
amogh thanks a lot

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