Question

`int_0^9 1/(1 + sqrtx)` dx = ______ 

Options

• 4 - 2 log 2 

• 4 + 2 log 2

• 6 - 4 log 2

• 6 + 4 log 2

Answer 1

`int_0^9 1/(1 + sqrtx)` dx = 6 - 4 log 2

Explanation:

Let I = `int_0^9 1/(1 + sqrtx)` dx

Put `1 + sqrtx = t`

⇒ x = (t - 1)²

⇒ dx = 2(t - 1)dt

∴ 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 (3 - log 2²)

∴ I = 6 - 4 log 2