What is the integral from-infinity to-1 of 1/x ?

The integral from-infinity to-1 of 1/x is diverges

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integral from-infinity to-1 of 1/x

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

diverges

Solution steps

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

Compute the indefinite integral:

\int \frac{1}{x} d x = ln ∣ x ∣ + C

Compute the boundaries:

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

diverges

= diverges

\frac{d}{d x} (\frac{x ^{3} + 3}{x})

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

\int 7 x^{\frac{2}{5}} + 4 x^{- \frac{4}{5}} d x

\int 6 x 1 + 3^{2} d x

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

What is the integral from-infinity to-1 of 1/x ?
The integral from-infinity to-1 of 1/x is diverges