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

The integral from 1 to e^2 of ((In(x))^2)/x is ((e^4-1)I^2n^2)/2

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integral from 1 to e^2 of ((In(x))^2)/x

\int_{1}^{e^{2}} \frac{( I n ( x ) ) ^{2}}{x} d x

\frac{( e ^{4} - 1 ) I ^{2} n ^{2}}{2}

Solution steps

\int_{1}^{e^{2}} \frac{( I n x ) ^{2}}{x} d x

Simplify

\frac{( I n x ) ^{2}}{x} : I^{2} n^{2} x

= \int_{1}^{e^{2}} I^{2} n^{2} x d x

Take the constant out:

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

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

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{1}^{e^{2}}

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

Compute the boundaries:

\frac{e ^{4}}{2} - \frac{1}{2}

= I^{2} n^{2} (\frac{e ^{4}}{2} - \frac{1}{2})

Simplify

= \frac{( e ^{4} - 1 ) I ^{2} n ^{2}}{2}

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What is the integral from 1 to e^2 of ((In(x))^2)/x ?
The integral from 1 to e^2 of ((In(x))^2)/x is ((e^4-1)I^2n^2)/2