# What is integral int 1/(x^2 - 3x + 2) dx

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### 2 Answers

The integral `int 1/(x^2 - 3x + 2) dx` has to be determined

`1/(x^2 - 3x + 2)`

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

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

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

Let `1/((x - 1)(x - 2)) = A/(x - 1) + B/(x - 2)`

=> `Ax - 2A + Bx - B = 1`

=> A + B = 0 and 2A + B = -1

=> A = -1, B = 1

`int 1/(x^2 - 3x + 2)`

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

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

= `ln(x - 2) - ln(x - 1) + C`

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

**The integral **`int 1/(x^2 - 3x + 2) dx = ln((x - 2)/(x - 1)) + C`

Sorry, it is int 1/(x^2 - 3x + 2) dx