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