Question

`int_0^4 1/(1 + sqrtx)`dx = ______.

Options

• `log (e^4/6)`

• `log (e^4/3)`

• `log (e^4/9)`

• `log (e^4/4)`

Answer 1

`int_0^4 1/(1 + sqrtx)`dx = `underlinebb(log (e^4/9))`

Explanation:

Let I = `int_0^4 1/(1 + sqrtx)`dx

Putting, `1 + sqrtx` = t

`=> 1/(2sqrtx) "dx" = "dt"`

`=> "dx" = 2sqrtx "dt" => "dx" = 2("t" - 1)"dt"`

at x = 0, t = 1 and x = 4, t = 3

Now, 

I = `int_1^3 (2("t" - 1)"dt")/"t"`

`= 2 int_1^3 (1 - 1/"t") "dt"`

`= 2["t" - log |"t"|]_1^3`

= 2[(3 - log 3) - (1 - log 1)]

= 2[2 - log 3]   ....(∵ log 1 = 0)

`= 4 + 2log (1/3) = 4 log e + log (1/9)`

`= log e^4 + log (1/9)`

`= log (e^4/9)`