Question

How do you evaluate `int 1/sqrt(2x+1)` for [3, 4]?

Answer 1

`3-sqrt7`

Explanation:

`I = int_(3)^(4) 1/sqrt(2x+1) dx`

`2x+1=t => 2dx=dt => dx=dt/2`

`t_1 = 2*3+1=7`
`t_2 = 2*4+1=9`

`I = int_7^9 1/sqrtt dt/2 = 1/2 int_7^9 t^(-1/2) dt = 1/2 t^(1/2)/(1/2)=t^(1/2)`

`I = sqrt9-sqrt7 = 3-sqrt7`