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

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

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integral from y to x of 1/(x^2)

\int_{y}^{x} \frac{1}{x ^{2}} d x

- \frac{1}{x} + \frac{1}{y}

Solution steps

\int_{y}^{x} \frac{1}{x ^{2}} d x

Apply the Power Rule

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

= [- \frac{1}{x}]_{y}^{x}

Compute the boundaries:

- \frac{1}{x} + \frac{1}{y}

= - \frac{1}{x} + \frac{1}{y}

\int_{0}^{3} \frac{arctan ( x )}{2} d x

\int_{1}^{2} (1 - \frac{x}{2}) d x

\int_{- c}^{c} (c - x^{2}) - (x^{2} - c) d x

\int_{0}^{3} x ln (x) d x

\int_{0}^{2} \frac{x ^{2}}{x ^{3} + 4} d x

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