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\[f(x) = x/(x ^{2} + 1)^{2}\]
that appears to be an arctan
x^2 + 1 = 2x we need a 2 up top to ln(x) it

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multiply by 1; or rather 2/2 keep the top but pull out the 1/2
that might not be good tho :)
x ---------- (x^2 +1)^2 tan the bottom :) maybe
*alternate method time* 1/2 * f 2x/(x^4+1) u=x^2 du=2x a=1 arctan(x^2)+C
tan(t) = x tan^2(t) = x^2 tan^2 + 1 = sec^2 sec^2^2 = sec^4 tan(t) sec^4(t)
here is the rule I used:
tan(t) sec^-4(t) maybe :)
my way you would still have to convert dx to dt...
\[\int\limits_{ }^{ }1/ (a^2+u^2) = 1/a \arctan(u/a)+C\]
well, that equation looks terrible...
wow lol
tan(t) = x dt sec^2 = dx thats helpful to me i think
[S] tan(t) sec^-2(t) dt ....maybe
tan = sin/cos = sin sec-1..... mines just messy lol
Isn't that straight u substitution? f(x) = x (x^2 + 1) ^ -2 Let u= x^2 + 1 and du = 2x
[S] 1/2u^2 du ?
That would give you...\[^{} \int\limits_{}^{} 1/2u ^{2}du\]
u^-2 = -u^-1
I mean -2 as the expontent
-1/2u = -1/2(x^2+1) +C
\[1/2\int\limits_{}^{}u ^{-2}du\]
=-1/(2u) + C, then substitue x^2 + 1 back in for u

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