<s>[INST] U-substitution of 2x in trigonometric substitution Find
$$\int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}} \,\mathrm{d}x.$$
The text says to use substitution of $u = 2x$.  How did they get $u = 2x$ and not $u = x^3$?
 [/INST] The hint:
$$\frac{x^3}{\sqrt{(4x^2+9)^3}}=\frac{x^3+\frac{9}{4}x-\frac{9}{4}x}{\sqrt{(4x^2+9)^3}}=\frac{x}{4\sqrt{4x^2+9}}-\frac{9x}{4\sqrt{(4x^2+9)^3}}.$$
Can you end it now?
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