- #1
Briggs
- 34
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I have a question in my book to evaluate
[tex]\int_{2}^{3}_\frac{x^2-2x+3}{(x-1)^2}dx[/tex]
I have completed the square on the top line of the fraction so that
[tex]\frac{x^2-2x+3}{(x-1)^2}[/tex] = [tex]\frac{(x-1)^2+2}{(x-1)^2}[/tex]
I have simplified so I end up with [tex]\int_{2}^{3}1+\frac{2}{(x-1)^2}dx[/tex] and now I don't know how to integrate this function. I have never encountered an integral with a number other than one on top of the fraction.
[tex]\int_{2}^{3}_\frac{x^2-2x+3}{(x-1)^2}dx[/tex]
I have completed the square on the top line of the fraction so that
[tex]\frac{x^2-2x+3}{(x-1)^2}[/tex] = [tex]\frac{(x-1)^2+2}{(x-1)^2}[/tex]
I have simplified so I end up with [tex]\int_{2}^{3}1+\frac{2}{(x-1)^2}dx[/tex] and now I don't know how to integrate this function. I have never encountered an integral with a number other than one on top of the fraction.
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