Question

`int_0^1 1/(2x + 5)dx` = ______

Options

• `1/2 log (7/5)`

• `1/2 log(5/7)`

• `log (7/5)`

• `1/4 log(7/5)`

Answer 1

`1/2 log (7/5)`

Explanation:

Let I = `int_0^1 dx/(2x + 5)`

Put 2x + 5 = t
∴ 2dx = dt
∴ dx = `"dt"/(2)` 
When x = 0, t = 2(0) + 5 = 5
When x = 1, t = 2(1) + 5 = 7

∴ I = `(1)/(2) int_5^7 "dt"/"t"`

= `(1)/(2)[log|"t"|]_5^7`

= `(1)/(2)(log 7 - log 5)`

= `(1)/(2)log(7/5)`