Question

Find the integral, using techniques from this or the previous chapter.
∫x/√16+8x^2 dx.

Answer 1

`\displaystyle\int\frac x{√(16+8x^2)}\,\mathrm dx`

Let
`u=16+8x^2\implies\mathrm du=16x\,\mathrm dx`. Then the integral becomes

`\displaystyle\frac1{16}\int(\mathrm du)/(\sqrt u)\,\mathrm du=\frac18\sqrt u+C`

`=\frac18√(16+8x^2)+C`