Find integral need help calculus

Hi can someone help with steps :
Find integral :
(7x) / (3 + x^2) ^(4)
Thanks!

Use substitution:

u = (3 + x^2)
du = 2x dx

∫ (7x) / (3 + x^2) ^(4) dx
∫ (7/2) du / u^4 = 7/2 (-1/3 u^-3) + C
= -7/(6u^3) + C
= -7/(6(3+x^2)^3) + C

Thanks so much it means a lot but I am confused on where-7/(6u^3) comes from ?

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∫ 1/u^4 = (-1/3) (1/u^3) * 7/2 = -7/(6u^3)

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