# What is the antiderivative of y/(x-1) if y=x^2-3x+2?

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We have to find the anti-derivative of y/(x-1) given that y=x^2-3x+2

y/(x-1)

=> (x^2 - 3x + 2)/ (x - 1)

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

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

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

=> (x - 2)

The anti derivative of (x - 2)

= Int[(x - 2) dx]

=> Int [x dx] - Int[2 dx]

=> x^2/2 - 2x + C

**The required anti derivative of y/(x-1) is x^2/2 - 2x + C**

To calculate the antiderivative of the function, we'll have to determine the indefinite integral of (x^2 - 3x + 2)/(x-1)

We notice that the roots of the numerator are 1 and 2, therefore we can re-write the quadratic as a product of linear factors:

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

We'll calculate the indefinite integral of the given function:

Int (x^2 - 3x + 2)dx/(x-1) = Int (x-1)(x-2)dx/(x-1)

We'll simplify and we'll get:

Int (x-1)(x-2)dx/(x-1) = Int (x-2)dx = Int xdx - 2Int dx

Int (x^2 - 3x + 2)dx/(x-1) = x^2/2 - 2x + C

**The antiderivative of the function is Int (x^2 - 3x + 2)dx/(x-1) = x^2/2 - 2x + C.**