When you set [tex]u=4x^2+4[/tex], you end up with the differentials [tex]\mathrm du=8x\,\mathrm dx[/tex]. Multiplying both sides by 2 gives [tex]2\,\mathrm du=16x\,\mathrm dx[/tex].
Then the integral is
[tex]\displaystyle\int_0^1\frac{16x}{(4x^2+4)^2}\,\mathrm dx=\int_4^8\frac2{u^2}\,\mathrm du=-\frac2u\bigg|_{u=4}^{u=8}=-2\left(\frac18-\frac14\right)=\frac14[/tex]
