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

The integral from-infinity to-1 of 1/(x^2) is 1

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

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

1

Solution steps

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

Compute the indefinite integral:

\int \frac{1}{x ^{2}} d x = - \frac{1}{x} + C

Compute the boundaries:

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

= 1 - 0

Simplify

= 1

\int \frac{x ^{2} + 1}{x} \cdot ln (x) d x

\frac{3}{2 x + 4} + 9 x^{2}

x \to 0 + lim ((1 - 2 x)^{\frac{3}{x}})

\int \frac{x ( x + 1 )}{4 x ^{4}} d x

\frac{\partial}{\partial x} (y ln (x + y))

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