What is the integral from 0 to t of 1/x ?

The integral from 0 to t of 1/x is diverges

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integral from 0 to t of 1/x

\int_{0}^{t} \frac{1}{x} d x

diverges

Solution steps

\int_{0}^{t} \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= [ln ∣ x ∣]_{0}^{t}

Compute the boundaries:

diverges

= diverges

\int \frac{x ^{3} - 6 x + 5}{x} d x

t an g e n t f (x) = 1 - 3 x^{[2]}, at x = 0

\int \frac{x}{3 x ^{2} - 4} d x

\frac{\partial}{\partial x} (ln (3 + x^{2} y^{2}))

\frac{d}{d x} (sin (e^{5 x}))

What is the integral from 0 to t of 1/x ?
The integral from 0 to t of 1/x is diverges