Question

求定积分，两道题

Answer 1

1. 因为 1/x - 1/(x+1) = [(x+1) - x]/[x(x+1)] = 1/[x(x+1)]

   所以，该积分就可以转换成：

   =∫dx/x - ∫dx/(x+1)

   =∫dx/x - ∫d(x+1)/(x+1)          注：d(x+1) = dx

   =[lnx - ln(x+1)]|x=1 →3

   =[(ln3 - ln1) - (ln4 - ln2)]

   =ln3 - (2ln2 - ln2)

   =ln3 - ln2

   =ln(3/2)

2. 设 x = 2sinα。那么就有，dx = 2cosα*dα。α∈[0, π/6] 注：当x = 1时，α=π/6

   那么，分母 √(4-x²) = √(4-4sin²α) = 2cosα

   因此，这个积分就可以变换为：

   =∫(2cosα*dα)/(2cosα)

   =∫dα

   =α|α=0 →π/6

   =π/6

Answer 2

Answer 3

Answer 4