What is the integral from 0 to 2 of 1/(x^2-5x+4) ?

The integral from 0 to 2 of 1/(x^2-5x+4) is diverges

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integral from 0 to 2 of 1/(x^2-5x+4)

\int_{0}^{2} \frac{1}{x ^{2} - 5 x + 4} d x

diverges

Solution steps

\int_{0}^{2} \frac{1}{x ^{2} - 5 x + 4} d x

Find undefined (singularity) points:

x = 1

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{1} \frac{1}{x ^{2} - 5 x + 4} d x + \int_{1}^{2} \frac{1}{x ^{2} - 5 x + 4} d x

\int_{0}^{1} \frac{1}{x ^{2} - 5 x + 4} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d}{d x} (\frac{2}{1 + e ^{- x}})

n = 0 \sum \infty 3 (- \frac{1}{2})^{n}

x \to 2 lim (\frac{4}{( x - 2 ) ^{5}})

f (x) = x + e^{x} - \frac{1}{x}, at x = 1

\int - \frac{2 x}{x ^{4} - 1} d x

What is the integral from 0 to 2 of 1/(x^2-5x+4) ?
The integral from 0 to 2 of 1/(x^2-5x+4) is diverges