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

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

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integral from 0 to e of (1+ln(x))/x

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

diverges

Solution steps

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

Expand

\frac{1 + ln ( x )}{x} : \frac{1}{x} + \frac{ln ( x )}{x}

= \int_{0}^{e} \frac{1}{x} + \frac{ln ( x )}{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}^{e} \frac{1}{x} d x + \int_{0}^{e} \frac{ln ( x )}{x} d x

\int_{0}^{e} \frac{1}{x} 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 e of (1+ln(x))/x ?
The integral from 0 to e of (1+ln(x))/x is diverges