# What is primitive function of `f(x)=1/(x^2+3x+2)` ?

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### 1 Answer

You need to find the primitive function of `f(x)` , hence, you need to evaluate the indefinite integral of the given function, such that:

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

You need to convert the denominator into its factored form and then you should use partial fraction expansion, such that:

`x^2+3x+2 = (x - x_1)(x - x_2)`

You should evaluate the roots `x_(1,2)` , using quadratic formula, such that:

`x_(1,2) = (-3+-sqrt(9 - 8))/2`

`x_(1,2) = (-3+-1)/2 => x_1 = -1 ; x_2 = -2`

`x^2+3x+2 = (x + 1)(x + 2)`

You need to use partial fraction expansion, such that:

`1/(x^2+3x+2) = 1/((x + 1)(x + 2)) = a/(x+1) + b/(x+2)`

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

`1 = ax + bx + 2a + b`

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

Equating the coefficients of like powers, yields:

`{(a + b = 0),(2a + b = 1):} => {(a = -b),(-2b + b = 1):} `

`{(a = -b),(-b = 1):}=> {(a = 1),(b = -1):} `

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

Intgerating both sides yields:

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

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

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

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

Using the properties of logarithms yields:

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

**Hence, evaluating the primitive function of `f(x)` , yields **`int 1/(x^2+3x+2) dx = ln|(x + 1)/(x + 2)| + c.`