# What is antiderivative of function y=1/(`sqrt(x+1)` +`sqrt x` )?

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the anti-derivative of the given function, hence, you need to evaluate the following indefinite integral, such that:

`int ydx = int 1/(sqrt(x+1) + sqrt x)dx`

You should perform the multiplication by the conjugate of denominator, such that:

`1/(sqrt(x+1) + sqrt x) = (sqrt(x+1) - sqrt x)/((sqrt(x+1) + sqrt x)(sqrt(x+1) - sqrt x))`

Converting the special product to denominator into a difference of squares, yields:

`(sqrt(x+1) - sqrt x)/(x + 1 - x) = sqrt(x+1) - sqrt x`

Replacing `sqrt(x+1) - sqrt x` for `1/(sqrt(x + 1) + sqrt x)` yields:

`int 1/(sqrt(x+1) + sqrt x) dx = int (sqrt(x+1) - sqrt x) dx`

Using the property of linearity of integral, yields:

`int 1/(sqrt(x+1) + sqrt x) dx = int sqrt(x+1)dx - int sqrt x dx`

You need to use the following substitution, such that:

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

Replacing the variable in `int sqrt(x+1)dx` yields:

`int sqrt(x+1)dx = int sqrt t dt`

`int sqrt t dt = int t^(1/2)dt = t^(1/2+1)/(1/2+1) + c`

Replacing back `x + 1` for t yields:

`int sqrt(x+1)dx = (2/3)(x+1)sqrt(x+1) + c`

`int sqrt x dx = (2/3)x*sqrt x + c`

`int 1/(sqrt(x+1) + sqrt x) dx = (2/3)(x+1)sqrt(x+1) - (2/3)x*sqrt x + c`

Factoring out `2/3` yields:

`int 1/(sqrt(x+1) + sqrt x) dx = (2/3)((x+1)sqrt(x+1) - x*sqrt x) + c`

Hence, evaluating the anti-derivative of the given function, yields` int 1/(sqrt(x+1) + sqrt x) dx = (2/3)((x+1)sqrt(x+1) - x*sqrt x) + c.`