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

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

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integral from-1 to 1 of (x+1)/(x-1)

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

diverges

Solution steps

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

\frac{x + 1}{x - 1} = \frac{2}{x - 1} + 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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