What is the integral from 0 to x of 1/(x^2) ?

The integral from 0 to x of 1/(x^2) is diverges

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integral from 0 to x of 1/(x^2)

\int_{0}^{x} \frac{1}{x ^{2}} d x

diverges

Solution steps

\int_{0}^{x} \frac{1}{x ^{2}} d x

Apply the Power Rule

: [- \frac{1}{x}]_{0}^{x}

= [- \frac{1}{x}]_{0}^{x}

Compute the boundaries:

diverges

= diverges

\frac{d}{d x} (sin^{4} (3 x))

\int r^{2} (\frac{r ^{3}}{24} - 6)^{7} d r

sin (t + 2)

f (x) = \frac{6}{x - 1}, (3, 3)

y = 6 x^{2} - 7 x + 4, (1, 3)

What is the integral from 0 to x of 1/(x^2) ?
The integral from 0 to x of 1/(x^2) is diverges