Question

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


`\int\limits^1_0 { (16x)/((4x^2+4)^2) } \, dx`

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(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`