Calculus Question need some help with limit

evaluate the limit (0, 1) 6x/square root of 16+3x^2 dx.

Really need explanation/step by step.

Thanks in advance!

1
∫ 6x dx / ( 3 x ² + 16 )^(1/2)
0

Let u = 3 x ² + 16
du = 6x dx

19
∫ du / u^(1/2)
16

19
∫ u^(-1/2) du
16

[2 u^(1/2) ] ----------limits 16 to 19

2 [ (19)^(1/2) - 4 ]

0.718

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