# How integrate y = 1/(x^2-7x+10)?

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### 1 Answer

You need to convert the quadratic denominator into its factored form, hence, you need to evaluate its solutions using the quadratic formula, such that:

`x^2 - 7x + 10 = 0`

`x_(1,2) = (7+-sqrt(49 - 40))/2 => x_(1,2) = (7+-sqrt9)/2`

`x_(1,2) = (7+-3)/2 => x_1 = 5, x_2 = 2`

Converting the quadratic denominator into its factored form, yields:

`x^2 - 7x + 10 = (x - 5)(x - 2)`

You need to integrate the following function, such that:

`int 1/(x^2 - 7x + 10)dx = int 1/((x - 5)(x - 2))dx`

You should use partial fraction expansion to split the integral into two simpler integrals, such that:

`1/((x - 5)(x - 2)) = a/(x - 5) + b/(x - 2)`

`1 = a(x - 2) + b(x - 5)`

`1 = ax - 2a + bx - 5b`

`1 = x(a + b) - 2a - 5b`

Equating the coefficients of like powers yields:

`{(a + b = 0),(-2a - 5b = 1):}`

`{(a + b = 0),(2b - 5b = 1):} => {(a + b = 0),(b = -1/3):} => {(a = 1/3),(b = -1/3):}`

`1/((x - 5)(x - 2)) = (1/3)(1/(x - 5) - 1/(x - 2))`

Integrating both sides, yields:

`int 1/((x - 5)(x - 2)) d x= (1/3)(int 1/(x - 5) dx - int 1/(x - 2) dx)`

`int 1/((x - 5)(x - 2)) dx = (1/3)(ln|x - 5| - ln|x - 2|) + c`

Using logarithmic identities yields:

`int 1/((x - 5)(x - 2)) dx = (1/3)(ln|(x - 5)/(x - 2)|) + c`

**Hence, evaluating the indefinite integral, using partial fraction expansion, yields**` int 1/(x^2 - 7x + 10)dx = (1/3)(ln|(x - 5)/(x - 2)|) + c.`