let f(x) = 3/ (x^1/2)

We need to find the integral of f(x).

==> intg f(x) = intg ( 3/(x^1/2) dx

We know that if 1/x^a = x^-a

==> intg f(x) = intg 3x^-1/2 dx

= 3intg x^-1/2 dx

= 3*x^1/2 / (1/2) + C

= 6x^1/2 + C

**==> intg 3/x^1/2 = 6x^1/2 + C**

We have to find the integral of 3/ sqrt x

Now Int [ 3/ x^(1/2)]

=> Int [3*x^(-1/2)]

=> 3* (x^(-1/2 +1))/(-1/2 +1) + C

=> 3* (x^1/2)/(1/2) + C

=> 6* sqrt x + C

**Therefore the integral of 3/(x^1/2) is 6*x^(1/2) + C**

To evaluate the integral of 3/(x^1/2).

Let f(x) = Int {3/(x^1/2) dx.

We use Int k*x^n dx = k*{ x^(n+1)}/(n+1) +C.

Therefore Int f(x) dx = Int 3 * x^ (-1/2) dx+ C.

Int f(x) dx = 3 * {x^(-1/2+1)}/(-1/2+1) +C.

Int f(x) dx = 3 {x^(1/2)}/(1/2) + C

Int f(x) dx = 6x^(1/2) + C.

We'll write the indefinite integral:

Int 3dx/sqrt x = 3Int dx/sqrt x

We'll multiply both, denominator and numerator, by 2:

2/2sqrtx = 2*(1/2sqrt x) = 2*(sqrtx)'

3Int dx/sqrt x = 3Int 2*(sqrtx)'dx

3Int 2*(sqrtx)'dx = 3*2Int (sqrtx)'dx

3*2Int (sqrtx)'dx = 6 sqrt x + C

**Int 3dx/sqrt x = 6 sqrt x + C**