What is the integral from 0 to infinity of |x^2-1| ?

The integral from 0 to infinity of |x^2-1| is diverges

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integral from 0 to infinity of |x^2-1|

\int_{0}^{\infty} x^{2} - 1 d x

diverges

Solution steps

\int_{0}^{\infty} x^{2} - 1 d x

Eliminate Absolutes

= \int_{0}^{1} - x^{2} + 1 d x + \int_{1}^{\infty} x^{2} - 1 d x

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

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

\int_{1}^{\infty} x^{2} - 1 d x =

diverges

= diverges

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What is the integral from 0 to infinity of |x^2-1| ?
The integral from 0 to infinity of |x^2-1| is diverges