# What are primitives of function f(x)=x^2/2-3x+2/(x-3)^3?

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### 1 Answer

You need to integrate the given function `f(x)` to evaluate its primitives, such that:

`int f(x)dx = F(x) + c`

`c in R` ( c represents the set that comprises all real constants)

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

Using the property of linearity of the integral, you may split the integral in three less difficult integrals, such that:

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

`int f(x)dx = (1/2)(x^3/3) - 3*x^2/2 + 2 int (x - 3)^(-3)dx`

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

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

**Hence, evaluating the primitives of the given function, using the process of integration, yields `F(x) = (1/2)(x^3/3) - 3*x^2/2 - 1/(x - 3)^2 + c` .**

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