What is the integral from 0 to 2 of (16)/(21x^2) ?

The integral from 0 to 2 of (16)/(21x^2) is diverges

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integral from 0 to 2 of (16)/(21x^2)

\int_{0}^{2} \frac{16}{21 x ^{2}} d x

diverges

Solution steps

\int_{0}^{2} \frac{16}{21 x ^{2}} d x

Take the constant out:

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

= \frac{16}{21} \cdot \int_{0}^{2} \frac{1}{x ^{2}} d x

Apply the Power Rule

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

= \frac{16}{21} [- \frac{1}{x}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

9 x^{2}, at x = 2

\int \frac{x}{x ^{2} - x + 6} d x

\frac{\partial}{\partial x} (\frac{a}{x ^{2}})

\frac{\partial}{\partial x} (ln (x^{6} y))

\frac{d y}{d x} + \frac{y}{x} = 8 x^{5} y^{2}

What is the integral from 0 to 2 of (16)/(21x^2) ?
The integral from 0 to 2 of (16)/(21x^2) is diverges