What is the integral from 1959 to infinity of 1/x ?

The integral from 1959 to infinity of 1/x is diverges

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integral from 1959 to infinity of 1/x

\int_{1959}^{\infty} \frac{1}{x} d x

diverges

Solution steps

\int_{1959}^{\infty} \frac{1}{x} d x

Compute the indefinite integral:

\int \frac{1}{x} d x = ln ∣ x ∣ + C

Compute the boundaries:

\int_{1959}^{\infty} \frac{1}{x} d x =

diverges

= diverges

\int_{1}^{2} \frac{x}{( x ^{2} + 3 ) ^{3}} d x

\int_{0.6}^{1} 3 x^{2} d x

\int_{0}^{2} π (4 - x^{2}) d x

\int_{1}^{3} \frac{π}{y} d y

\int_{0}^{l n (4)} e^{3 x} d x

What is the integral from 1959 to infinity of 1/x ?
The integral from 1959 to infinity of 1/x is diverges