How to evaluate the definite integral of 1/(1+square root x), for limis of integration 0 and 4?

First, we'll determine the indefinite integral of the
function y. For this reason, we'll replace the variable x by
t^2.

x = t^2

We'll
differentiate both sides:

dx =
2tdt

Int  dx/(sqrt x + 1) = Int [1/(t +1)]*(2t
dt)

Int [1/(t +1)]*(2t dt) = 2Int (t+1)dt/(t +1) - 2Int
dt/(t+1)

Int [1/(t +1)]*(2t dt) = 2Int dt - 2Int
dt/(t+1)

We'll calculate the 1st
term:

2Int dt = 2t + C

We'll
calculate the 2nd term:

2Int dt/(t+1) = 2 ln|t+1| +
C

Int  dx/(sqrt x + 1) = 2sqrt x - 2 ln |sqrt x + 1| +
C

We'll apply Leibniz Newton to determine the definite
integral:

Int  dx/(sqrt x + 1) = F(4) -
F(0)

F(4) = 2sqrt 4 - 2 ln |sqrt 4 +
1|

F(4) = 4 - 2 ln 3

F(4) = 4
- ln 9

F(0) = 2sqrt 0 - 2 ln |sqrt 0+
1|

F(0) = 0 - ln 1

F(0) =
0

The definite integral of the given
function, if the limits of integration are x = 0 to x = 4,
is: Int dx/(sqrt x + 1) = 4 - ln
9