1.
∫_(-1)^(2) 1/(11 + 5x)³ dx
= (1/5)∫_(-1)^(2) 1/(11 + 5x)³ d(5x)
= (1/5)∫_(-1)^(2) (11 + 5x)^(-3) d(11 + 5x)
= (1/5) (11 + 5x)^(-3 + 1)/(-3 + 1)_(-1)^(2)
= (1/5)(-1/2)[(11 + 5*2)^(-2) - (11 + 5*-1)^(-2)]
= 1/392
2.
∫_(1)^(4) 1/(1 + √x) dx
令u = √x,u² = x,2udu = dx
当x = 1,u = 1；当x = 4,u = 2
∫_(1)^(2) 1/(1 + u) * 2udu
= 2∫_(1)^(2) (u + 1 - 1)/(1 + u) du
= 2∫_(1)^(2) [1 - 1/(1 + u)] du
= 2[u - ln|1 + u|]_(1)^(2)
= 2[(2 - ln3) - (1 - ln2)]
= 2 + ln(4/9)