# find the antiderivative of a square root functionFind the antiderivative of y= square root(1+square rootx).

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You need to consider the inverse operation to differentiation, such that:

`int sqrt(1 + sqrt x)dx`

You need to come up with the following substitution, such that:

`sqrt x = u => (1/(2sqrt x)) dx = du => dx = 1/(1/(2u)) du`

`dx = 2u du`

Changing the variable yields:

`2int u*sqrt(1 + u) du`

You should come up with the next substitution, such that:

`1 + u = v => u = v - 1`

`du = dv`

Changing again the variable yields:

`int (v - 1)sqrt v dv = int (vsqrt v - sqrt v)dv`

Using the property of linearity of integral yields:

`int (vsqrt v - sqrt v)dv = int (vsqrt v) dv - int (sqrt v) dv`

Converting the square root into a rational power, yields:

`int (vsqrt v - sqrt v)dv = int (v*v^(1/2)) dv - int (v^(1/2)) dv`

`int (vsqrt v - sqrt v)dv = int (v^(1/2 + 1)) dv - int (v^(1/2)) dv`

`int (vsqrt v - sqrt v)dv = int (v^(3/2)) dv - int (v^(1/2)) dv`

`int (vsqrt v - sqrt v)dv = (v^(3/2 + 1))/(3/2 + 1) - (v^(1/2 + 1))/(1/2 + 1) + c`

`int (vsqrt v - sqrt v)dv = (2/5)sqrt(v^5) - (2/3)sqrt(v^3) + c`

`int (vsqrt v - sqrt v)dv = (2/5)v^2*sqrt v - (2/3)v*sqrt v + c`

Substituting back `1 + sqrt x` for v yields:

`int sqrt(1 + sqrt x)dx = (2/5)(1 + sqrt x)^2*sqrt(1 + sqrt x) - (2/3)(1 + sqrt x)*sqrt (1 + sqrt x) + c`

**Hence, evaluating the antiderivative of the given function yields **`int sqrt(1 + sqrt x)dx = (2/5)(1 + sqrt x)^2*sqrt(1 + sqrt x) - (2/3)(1 + sqrt x)*sqrt (1 + sqrt x) + c.`

To find the antiderivative of the square root function, we'll have to determine the indefinite integral.

Int sqrt(1+sqrt x) dx

We'll substitute 1 + sqrt x = t

We'll differentiate both sides:

dx/2sqrt x = dt

dx = 2 sqrt x*dt

But sqrt x = t - 1

dx = 2(t - 1)dt

We'll re-write the integral in t:

Int sqrt t*2(t - 1)dt = 2Int sqrt t^3 dt - 2 Int sqrt t dt

Int sqrt t*2(t - 1)dt = 2*t^(3/2 + 1)/(3/2 + 1) - 2*t^(1/2 + 1)/(1/2 + 1) + C

Int sqrt t*2(t - 1)dt = 4t^(5/2)/5 - 4t^(3/2)/3 + C

Int sqrt t*2(t - 1)dt = 4t^(3/2)( t/5 - 1/3) + C

**Int sqrt(1+sqrt x)dx = 4(1 + sqrt x)^(3/2)[(1 + sqrt x)/5 - 1/3] + C**

Antiderivate is the opposite of a derivative, that is to say, whatever the rules are for taking the derivative of a function are opposite in this case.

I will use x to be the antiderivate of y.

first, we need to change square root as a power of 1/2 to

step 1: x = 1 + sqrt(x) ^ 1/2

next, we bring it down and add 1 to the power.

step 2: x = 1 + sqrt(x) ^ 1/2 + 2/2 / 2

step 3: x = 1+sqrt(x)^3/2 / 2

Final answer: **Antiderivate of y is 1+sqrt(x)^3/2 over 2.**