How do I integrate this equation

I'm just beginning integration in University and need a little help with a question.

I need to integrate:

3t(t^2+8)^(1/2)

So 3t multiplied by the square root of (t^2+8)

A simple walkthrough would be very helpful! Thanks!

........... ∫ 3t√(t²+8) dt

= (3/2) ∫ √(t²+8) · 2t dt

= (3/2) ∫ √(u) du, ............ u = t²+8, du = 2t dt

= (3/2)· ( 2 √u ) + C

= 3·√(t²+8) + C ...................... Ans.
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