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

Compute the indefinite integral:

\int - \frac{1}{x} d x = - ln ∣ x ∣ + C

Compute the boundaries:

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

diverges

= diverges

(cos (3 x)) y^{^{'}} - 3 (tan (3 x)) y = - 10 sec (3 x)

(t + 1) cos (- t)

\int - 2 cos^{2} (x) sin (x) d x

\frac{d v}{d t} = 32 - \frac{4}{5} v

\frac{\partial}{\partial x} (x^{5} y^{5} + 4 x^{7} y)

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