Evaluate the indefinite integral?..?

Question

Integral ((3x)/(x^4+1))dx?

Answer 1

Let u = x^2, du = 2xdx.
Then the integral becomes
3/2 * ∫ du / (u²+1) = 3/2 * arctan u = (3/2) arctan(x²)+C.

Answer 2

The trick here is to change into one of the inverse trig identity form. for example integral1/x^2 + 1 = arctan(x) + c so what do we have to do in order to change 3x/(x^4 + 1) to the form off 1/x^2 + 1.

Well the first thing we can do is pull three out in the front. so we have
3 integral (x/x^4 + 1) next thing to do is rewrite x^4 as x^2^2 that gives us
3 integral (x/(x^2)^2 + 1)
now the next step is to do a substitution that is
let u = x^2 then
du = 2xdx
du/2 = xdx now with this the integral becomes
3/2 integral 1/(u^2 + 1) * du [notice this is in form 1/x^2 + 1of]
which equals
3/2 integral du/(u^2 + 1) now solving this integral is easy it is a inverse trig type so we get
[3/2][arctan(u)] + c now since this is an indefinite integral we have to substitute or value for u which was x^2 so the final answer is
[3/2][arctan(x^2)] + c

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