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

The integral from 0 to 1 of 1/(x(x+2)) is diverges

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integral from 0 to 1 of 1/(x(x+2))

\int_{0}^{1} \frac{1}{x ( x + 2 )} d x

diverges

Solution steps

\int_{0}^{1} \frac{1}{x ( x + 2 )} d x

Take the partial fraction of

\frac{1}{x ( x + 2 )} : \frac{1}{2 x} - \frac{1}{2 ( x + 2 )}

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d}{d x} (\frac{e ^{x}}{x ^{5}})

\frac{d}{d x} ((1 - \frac{x}{7})^{- 7})

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\frac{d}{d x} (9 arctan (x))

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