# What is the primitive of `f(x)=1/x(x^2+1)` ?

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### 1 Answer

You need to evaluate the primitive of the function` f(x) = 1/(x(x^2 + 1))` , hence, you need to evaluate the following indefinite integral, such that:

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

You should use the partial fraction expansion to decompose the fraction into two simpler fractions, such that:

`1/(x(x^2 + 1)) = a/x + (bx + c)/(x^2 + 1)`

`1 = a(x^2 + 1) + x*(bx + c)`

`1 = ax^2 + a + bx^2 + cx`

`1 = x^2(a + b) + cx + a`

Equating the coefficients of equal powers yields:

`a + b = 0 => a = -b`

`c = 0`

`a = 1 => b = -1`

`1/(x(x^2 + 1)) = 1/x - 1/(x^2 + 1)`

Integrating both sides, yields:

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

Using the property of linearity of integral, you need to split the integral into two simpler integrals, such that:

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

`int 1/(x(x^2 + 1)) dx = ln|x| - tan^(-1)(x) + c`

**Hence, evaluating the primitive of the given function, using the integration process, yields **`int 1/(x(x^2 + 1)) dx = = ln|x| - tan^(-1)(x) + c.`

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