∫ dx/[(x + 1)(1 + √(1 + x))]
令t² = x + 1,2tdt = dx,x∈[0,3] ==> t∈[1,2]
= ∫ 2t/[t²(1 + t)] dt
= 2∫ (1 + t - t)/[t(1 + t)] dt
= 2∫ [1/t - 1/(1 + t)] dt
= 2[ln|t| - ln|1 + t|]
= 2ln|t/(1 + t)|
= 2ln(2/3) - 2ln(1/2)
= 2ln(4/3)