Question

Can somebody explain this to me when 2du = 16xdx ???

Answer 1

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

$\displaystyle\int_0^1\frac{16x}{(4x^2+4)^2}\,\mathrm dx=\int_4^8\frac2{u^2}\,\mathrm du=-\frac2u\bigg|_{u=4}^{u=8}=-2\left(\frac18-\frac14\right)=\frac14$