Answer 1

Let I = `int_0^9 (1)/(1 + sqrt(x))*dx`
Put `1 + sqrt(x)` = t
∴ x = (t – 1)²
∴ dx = 2(t – 1)dt
When x = 0, t = 1 + 0 = 1
When x = 9, t = `1 + sqrt(9)`
= 1 + 3 = 4

∴ I = `int_1^4 (2(t - 1))/"t"*"dt"`

= `2int_1^4(1 - 1/"t")*"dt"`

= `2]"t" - log|"t"|]_1^4`
= 2 [(4 – log |4|) – (1 – log |1|)]
= 2 [4 – log 4 – (1 – 0)]
= 2 [4 – log 2² – 1)
= 2 (3 – 2log 2)
∴ I = 6 – 4 log 2.