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

1 Answer
Oct 15, 2015

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