complete the following integral or partial substitution with a. ∫(x^5+2x)/(x^6+6x^2+56)^2 dx

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complete the following integral or partial substitution with a. ∫(x^5+2x)/(x^6+6x^2+56)^2 dx

Mathematics
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∫(x^5+2x)/(x^6+6x^2+56)^2 dx ∫(x^5+2x) dx - ∫(x^6+6x^2+56)^-2 dx
You can't do this in products, only in additions. suzi20
u = x^6 +6x^2 +56 du = 6x^5 +12x dx du = 6(x^5 +2x) dx

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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
Suzi please continue the answer
u have to do partial again
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)
I could ask you to complete
that's calculus 2, i remember now, wow amogh cool...........
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
he did
Tell me what you didn't understand, I've written it completely!
oh I'm sorry, I really do not understand about the calculus lesson, but I want to learn
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)
amogh thanks a lot

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