# What is primitive of y=x/(3-x^2)^(1/2)?

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### 1 Answer

You need to evaluate the indefinite integral of the given function to find its primitive, such that:

`f(x) = int x/sqrt(3 - x^2) dx`

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

`3 - x^2 = y => -2xdx = dy => xdx = -(dy)/2`

Replacing the variable to integrand yields:

`int (-(dy)/2)/sqrt y = -(1/2)int (dy)/(y^(1/2))`

`-(1/2)int (dy)/(y^(1/2)) = -(1/2)int (y^(-1/2))(dy)`

`-(1/2)int (y^(-1/2))(dy) = -(1/2)*y^(-1/2+1)/(-1/2+1) + c`

`-(1/2)int (y^(-1/2))(dy) = -(1/2)*y^(1/2)/(1/2) + c`

Reducing duplicate factors yields:

`-(1/2)int (y^(-1/2))(dy) = -sqrt y + c`

Replacing back `3 - x^2` for `y` yields:

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

**Hence, evaluating the primitive f(x) of the given function, yields **`f(x) = -sqrt(3 - x^2) + c.`