Definite Intergration

a) Note that :

1/[u(u + 1)] = 1/u - 1/(u + 1)

So:

∫ du/[u(u + 1)] = ∫du/u - ∫du/(u + 1)

= ln u - ln (u + 1) + C

= ln [u/(u + 1)] + C

b) Sub u = ex, then du = exdx, i.e. dx = du/u

∫ (x = ln 2 → ln 4) dx/(ex - 1)

∫ (u = 2 → 4) du/[u(u - 1)]

= [ln [u/(u + 1)]] (u = 2 → 4)

= ln (4/5) - ln (2/3)

= ln (6/5)