What is the integral from 0 to 2 of 4/(x^2) ?

The integral from 0 to 2 of 4/(x^2) is diverges

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integral from 0 to 2 of 4/(x^2)

\int_{0}^{2} \frac{4}{x ^{2}} d x

diverges

Solution steps

\int_{0}^{2} \frac{4}{x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{0}^{2} \frac{1}{x ^{2}} d x

Apply the Power Rule

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

= 4 [- \frac{1}{x}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

\frac{d}{d s} (\frac{- s ^{2} + 1}{( s ^{2} + 1 ) ^{2}})

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

\frac{d}{d x} (2 x^{4 π})

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

\frac{d}{d y} (x^{y^{z}})

What is the integral from 0 to 2 of 4/(x^2) ?
The integral from 0 to 2 of 4/(x^2) is diverges