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

The integral from-1 to 1 of 1/(x^2-2x) is diverges

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integral from-1 to 1 of 1/(x^2-2x)

\int_{- 1}^{1} \frac{1}{x ^{2} - 2 x} d x

diverges

Solution steps

\int_{- 1}^{1} \frac{1}{x ^{2} - 2 x} d x

Find undefined (singularity) points:

x = 0

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 1 \sum \infty \frac{2}{n} - \frac{2}{n + 1}

\int \frac{x ^{2}}{( 1 - x ^{2} ) ^{\frac{5}{2}}} d x

\frac{e ^{- 2 s}}{s ^{2}}

x \to - 2 lim (8)

f (x) = \frac{168}{( x + 4 ) ^{5}}

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