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You are welcome! Glad I could help!

I, personally, would not use a trig substitution here. Use long division instead!

From division:

x^3/(x^2 + 4) = x - 4x/(x^2 + 4).

Then, we have:

∫ x^3/(x^2 + 4) dx

=> ∫ [x - 4x/(x^2 + 4)] dx

= ∫ x dx - 4 ∫ x/(x^2 + 4) dx

= (1/2)x^2 - 4 ∫ x/(x^2 + 4) dx.

For ∫ x/(x^2 + 4), let u = x^2 + 4 <==> du = 2x dx. Then, we have:

(1/2)x^2 - 4 ∫ x/(x^2 + 4) dx

= (1/2)x^2 - 2 ∫ 2x/(x^2 + 4) dx

= (1/2)x^2 - 2 ∫ 1/u du, by applying substitutions

= (1/2)x^2 - 2ln|u| + C

= (1/2)x^2 - 2ln|x^2 + 4| + C, since u = x^2 + 4

= (1/2)x^2 - 2ln(x^2 + 4) + C, since x^2 + 4 > 0 for all x.

I hope this helps!

I agree with Brian, I would just use longhand division first !

But actually, there is no trig substitution used in either solution presented here.... though it looks like you did attempt to solve it that way. better to try an "easier method " first, before resorting to a trig sub ... [ one of my least favorite methods also !!! ]

the + 4x - 4x "trick" is not easy to see in the first method, but longhand division is really easy here.

∫(x³)/(x² + 4) dx

∫(x³ + 4x - 4x)/(x² + 4) dx

∫x(x² + 4)/(x² + 4) dx - 4*∫x/(x² + 4) dx

x²/2 - 2*ln(x² + 4) + C