What is the integral from-1 to 0 of 1/(x^3) ?

The integral from-1 to 0 of 1/(x^3) is diverges

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integral from-1 to 0 of 1/(x^3)

\int_{- 1}^{0} \frac{1}{x ^{3}} d x

diverges

Solution steps

\int_{- 1}^{0} \frac{1}{x ^{3}} d x

Apply the Power Rule

: [- \frac{1}{2 x ^{2}}]_{- 1}^{0}

= [- \frac{1}{2 x ^{2}}]_{- 1}^{0}

Compute the boundaries:

diverges

= diverges

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{6}{x ^{4}}) (x + 8 x^{3}))

x \to 1 lim (\frac{0}{x ^{2} - 1})

\int (18 x + 5) d x

\int 2 e^{3 x} d x

3 x \cdot \frac{d y}{d x} - y = x

What is the integral from-1 to 0 of 1/(x^3) ?
The integral from-1 to 0 of 1/(x^3) is diverges