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

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### 1 Answer

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**