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

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

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integral from-1 to infinity of 1/x

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

diverges

Solution steps

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

Adjust integral based on undefined pointsThere is an undefined point within the boundaries at

: 0

= \int_{- 1}^{0} \frac{1}{x} d x + \int_{0}^{\infty} \frac{1}{x} d x

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

diverges

= diverges

\int_{0}^{4} (x^{3} + 2) d x

\int_{2}^{\infty} \frac{1}{y ^{2} - 1} d y

\int_{- 1}^{1} (sin (x))^{2} d x

\int_{0}^{4} \frac{( 2 - x )}{4 x - x ^{2}} d x

\int_{x^{2}}^{e^{x}} t^{2} ln (t) d t

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