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How do you find the indefinite integral of #int (x^2-3x+2)/(x-1) dx#?

Answer 1

Charles Anctil

The function is #y = 1/2x^2 - 2x + C#.

I would recommend factoring the numerator and the denominator to see if we can eliminate/simplify before integrating.

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

#=>int(x - 2)#

#=> 1/2x^2 - 2x+ C#

You can always verify your answer by differentiating. You will see the integration worked.

Hopefully this helps!

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Answer 2

Axel Alvarez

To find the indefinite integral of ( \int \frac{x^2 - 3x + 2}{x - 1} , dx ), you can use polynomial long division to divide ( x^2 - 3x + 2 ) by ( x - 1 ), which will give you the quotient ( x - 2 ) and a remainder of ( 0 ). Then integrate the resulting expression. The indefinite integral is:

[ \int \frac{x^2 - 3x + 2}{x - 1} , dx = \int (x - 2) , dx ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.