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how am I able to solve this definite integral when it goes from 0 to 2; I know how to solve it for example

$\int _3^5\frac{2x-1}{x^2-3x+2}dx$

but not when limit is approaching from a negative number or a zero...

$\int _0^2\frac{2x-1}{x^2-3x+2}dx$

$\int _{-1}^2\:\frac{2x-1}{\:x^2-3x+2}dx$

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    Not possible. The integral is divergent (x=1 is a root of the denominator, split the integral at x=1 and you arrive at divergent integrals) – imranfat Oct 20 '16 at 20:34
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    $$f(x)=\frac{2x-1}{x^2-3x+2}=\frac{3}{x-2}-\frac{1}{x-1}$$ has non-integrable singularities (simple poles) at $x=1$ and $x=2$. Besides that, the partial fraction decomposition provides a straightforward way to compute its antiderivative. – Jack D'Aurizio Oct 20 '16 at 20:36
  • the roots of denominator are $1$ and $2$, so to be definite, the integration interval must not contain these values. – hamam_Abdallah Oct 20 '16 at 20:39

3 Answers3

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Here you have that in $(0,2)$ the integral diverges, because in $2$ it behaves like the integral of $\frac 1 x$ (multiplied for a positive constant) in $0$ that is divergent.

Bargabbiati
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  • ok say i have $\int _{-2}^2:\frac{2x-1}{:x^2-3x+2}dx$ ; here i can split the limit from -2 to -1 then 1to 2? –  Oct 20 '16 at 20:40
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    Bargagnati non mi rubare il lavoro! @TemurBakhriddinov: you may split what you like, but the integral over $(1,2)$ is still divergent. – Jack D'Aurizio Oct 20 '16 at 20:43
  • Yes, you can, but generally if somewhere the integral diverges to $+\infty$ and in another point of the region where you are integrating it diverges to $-\infty$ you can't say anything about the integral in the whole region. – Bargabbiati Oct 20 '16 at 20:44
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$$f(x)=\frac{2x-1}{x^2-3x+2}=\frac{3}{x-2}-\frac{1}{x-1}$$ has non-integrable singularities (simple poles) at $x=1$ and $x=2$. Besides that, the partial fraction decomposition provides a straightforward way to compute its antiderivative.
It also allows us to state

$$ PV\int_{0}^{2}\frac{(2x-1)\,dx}{x^2-3x+2} = PV\int_{0}^{2}\frac{3\,dx}{x-2}=-\infty,$$

$$ \int_{3}^{5}\frac{(2x-1)\,dx}{x^2-3x+2} = 3\log 3-\log 2.$$

Jack D'Aurizio
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It is an improper integral, so you have to split it, for example like that:

$$ \begin{align*} \int_0^2\frac{2x-1}{x^2-3x+2}dx&= \int_0^2\frac{2x-1}{(x-2)(x-1)}dx\\ &=\int_0^1\frac{2x-1}{(x-2)(x-1)}dx+\int_1^{3/2}\frac{2x-1}{(x-2)(x-1)}dx +\int_{3/2}^2\frac{2x-1}{(x-2)(x-1)}dx\\ &=\lim_{b\to1^-}\int_0^b\frac{2x-1}{(x-2)(x-1)}dx +\lim_{a\to1^+}\int_a^{3/2}\frac{2x-1}{(x-2)(x-1)}dx\\ &\qquad+\lim_{a\to1^+}\int_a^{3/2}\frac{2x-1}{(x-2)(x-1)}dx +\lim_{b\to2^-}\int_{3/2}^{b}\frac{2x-1}{(x-2)(x-1)}dx\\ \end{align*} $$

zar
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  • how do you know where to split for partial fraction problems? I have a problem where i think it's already splitted: –  Oct 20 '16 at 21:33
  • $\int _{\frac{3}{2}}^3:\frac{x-1}{x^2-3x+2}dx $ (how do you get it in the form of a fraction) –  Oct 20 '16 at 21:33
  • unless its 1, and 2; and 3/2 is 1.5 (sorry for making these multiple comments –  Oct 20 '16 at 21:35
  • Split as you like, as long as the point you choose is inside your domain – zar Oct 20 '16 at 21:56