(5) 令 √(x+9) = u, 则 x = u^2-9, dx = 2udu
I = ∫[√(x+9)/x]dx = ∫2u^2du/(u^2-9) = 2∫(u^2-9+9)du/(u^2-9)
= 2∫du + 3∫[1/(u-3) - 1/(u+3)]du = 2u+3ln|(u-3)/(u+3)| + C
= 2√(x+9) + 3ln|[√(x+9)-3]/[√(x+9)+3]| + C
(6) 令 √(x+1) = u, 则 x = u^2-1, dx = 2udu
I = ∫dx/[x√(x+1)] = ∫2du/(u^2-1) = ∫[1/(u-1) - 1/(u+1)]du
= ln|(u-1)/(u+1)| + C = ln|[√(x+1)-1]/[√(x+1)+1]| + C
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