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

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

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

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

x^{2} (- \frac{1}{x} + x)

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

= x^{2} [- \frac{1}{y}]_{\frac{1}{x}}^{x}

Compute the boundaries:

- \frac{1}{x} + x

= x^{2} (- \frac{1}{x} + x)

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