What is the integral from 0 to pi of (2-x)/(x^2) ?

The integral from 0 to pi of (2-x)/(x^2) is diverges

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integral from 0 to pi of (2-x)/(x^2)

\int_{0}^{π} \frac{2 - x}{x ^{2}} d x

diverges

Solution steps

\int_{0}^{π} \frac{2 - x}{x ^{2}} d x

Expand

\frac{2 - x}{x ^{2}} : \frac{2}{x ^{2}} - \frac{1}{x}

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

\int_{0}^{π} \frac{2}{x ^{2}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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What is the integral from 0 to pi of (2-x)/(x^2) ?
The integral from 0 to pi of (2-x)/(x^2) is diverges