# ∫202x−1×2−3x+2dx\int _0^2\:\frac{2x-1}{\:x^2-3x+2}dx

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

∫532x−1×2−3x+2dx\int _3^5\frac{2x-1}{x^2-3x+2}dx

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

∫202x−1×2−3x+2dx\int _0^2\frac{2x-1}{x^2-3x+2}dx

∫2−12x−1×2−3x+2dx\int _{-1}^2\:\frac{2x-1}{\:x^2-3x+2}dx

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1

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 at 20:34

1

f(x)=2x−1×2−3x+2=3x−2−1x−1f(x)=\frac{2x-1}{x^2-3x+2}=\frac{3}{x-2}-\frac{1}{x-1} has non-integrable singularities (simple poles) at x=1x=1 and x=2x=2. Besides that, the partial fraction decomposition provides a straightforward way to compute its antiderivative.
– Jack D’Aurizio
Oct 20 at 20:36

the roots of denominator are 11 and 22, so to be definite, the integration interval must not contain these values.
– Abdallah Hammam
Oct 20 at 20:39

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3

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Here you have that in (0,2)(0,2) the integral diverges, because in 22 it behaves like the integral of 1x\frac 1 x (multiplied for a positive constant) in 00 that is divergent.

ok say i have ∫2−22x−1×2−3x+2dx\int _{-2}^2\:\frac{2x-1}{\:x^2-3x+2}dx ; here i can split the limit from -2 to -1 then 1to 2?
– Temur Bakhriddinov
Oct 20 at 20:40

2

Bargagnati non mi rubare il lavoro! @TemurBakhriddinov: you may split what you like, but the integral over (1,2)(1,2) is still divergent.
– Jack D’Aurizio
Oct 20 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.
– Giuseppe Bargagnati
Oct 20 at 20:44

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

∫202x−1×2−3x+2dx=∫202x−1(x−2)(x−1)dx=∫102x−1(x−2)(x−1)dx+∫3/212x−1(x−2)(x−1)dx+∫23/22x−1(x−2)(x−1)dx=limb→1−∫b02x−1(x−2)(x−1)dx+lima→1+∫3/2a2x−1(x−2)(x−1)dx+lima→1+∫3/2a2x−1(x−2)(x−1)dx+limb→2−∫b3/22x−1(x−2)(x−1)dx
\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\\
+\lim_{b\to2^-}\int_{3/2}^{b}\frac{2x-1}{(x-2)(x-1)}dx\\
\end{align*}

how do you know where to split for partial fraction problems? I have a problem where i think it’s already splitted:
– Temur Bakhriddinov
Oct 20 at 21:33

∫332x−1×2−3x+2dx\int _{\frac{3}{2}}^3\:\frac{x-1}{x^2-3x+2}dx (how do you get it in the form of a fraction)
– Temur Bakhriddinov
Oct 20 at 21:33

unless its 1, and 2; and 3/2 is 1.5 (sorry for making these multiple comments
– Temur Bakhriddinov
Oct 20 at 21:35

Split as you like, as long as the point you choose is inside your domain
– zar
Oct 20 at 21:56

f(x)=2x−1×2−3x+2=3x−2−1x−1f(x)=\frac{2x-1}{x^2-3x+2}=\frac{3}{x-2}-\frac{1}{x-1} has non-integrable singularities (simple poles) at x=1x=1 and x=2x=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, 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. \int_{3}^{5}\frac{(2x-1)\,dx}{x^2-3x+2} = 3\log 3-\log 2.