What is the integral from 0 to 2 of (3x^2-1)/(x^2) ?

The integral from 0 to 2 of (3x^2-1)/(x^2) is diverges

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integral from 0 to 2 of (3x^2-1)/(x^2)

\int_{0}^{2} \frac{3 x ^{2} - 1}{x ^{2}} d x

diverges

Solution steps

\int_{0}^{2} \frac{3 x ^{2} - 1}{x ^{2}} d x

Expand

\frac{3 x ^{2} - 1}{x ^{2}} : 3 - \frac{1}{x ^{2}}

= \int_{0}^{2} 3 - \frac{1}{x ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{2} 3 d x - \int_{0}^{2} \frac{1}{x ^{2}} d x

\int_{0}^{2} 3 d x = 6

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

y = 2 x^{3} - x^{2} + 7

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

\frac{e ^{x} + 1}{e ^{x} - 1}

\frac{d y}{d x} = 2 x y^{2} + 128 x

x \to - 6 lim (12 x + 1)

What is the integral from 0 to 2 of (3x^2-1)/(x^2) ?
The integral from 0 to 2 of (3x^2-1)/(x^2) is diverges