What is the integral from 9 to infinity of 1/x ?

The integral from 9 to infinity of 1/x is diverges

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integral from 9 to infinity of 1/x

\int_{9}^{\infty} \frac{1}{x} d x

diverges

Solution steps

\int_{9}^{\infty} \frac{1}{x} d x

Compute the indefinite integral:

\int \frac{1}{x} d x = ln ∣ x ∣ + C

Compute the boundaries:

\int_{9}^{\infty} \frac{1}{x} d x =

diverges

= diverges

\int_{0}^{6} e^{x} - e^{- x} d x

\int_{5}^{26} x e^{- x^{2}} d x

\int_{1}^{8} 1 + 4 x^{- \frac{2}{3}} d x

\int_{0}^{9 - y^{2}} z d z

\int_{0}^{1} 4 x^{3} (8 - x^{4}) d x

What is the integral from 9 to infinity of 1/x ?
The integral from 9 to infinity of 1/x is diverges