What is the integral from 1 to 10 of 1/(xIn^2x) ?

The integral from 1 to 10 of 1/(xIn^2x) is 9/(10In^2)

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integral from 1 to 10 of 1/(xIn^2x)

\int_{1}^{10} \frac{1}{x I n ^{2} x} d x

\frac{9}{10 I n ^{2}}

Solution steps

\int_{1}^{10} \frac{1}{x I n ^{2} x} d x

Take the constant out:

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

= \frac{1}{I n ^{2}} \cdot \int_{1}^{10} \frac{1}{xx} d x

xx = x^{2}

= \frac{1}{I n ^{2}} \cdot \int_{1}^{10} \frac{1}{x ^{2}} d x

Apply the Power Rule

: [- \frac{1}{x}]_{1}^{10}

= \frac{1}{I n ^{2}} [- \frac{1}{x}]_{1}^{10}

Simplify

\frac{1}{I n ^{2}} [- \frac{1}{x}]_{1}^{10} : \frac{[ - \frac{1}{x} ] _{1}^{10}}{I n ^{2}}

= \frac{[ - \frac{1}{x} ] _{1}^{10}}{I n ^{2}}

Compute the boundaries:

\frac{9}{10}

= \frac{\frac{9}{10}}{I n ^{2}}

Simplify

= \frac{9}{10 I n ^{2}}

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What is the integral from 1 to 10 of 1/(xIn^2x) ?
The integral from 1 to 10 of 1/(xIn^2x) is 9/(10In^2)