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^4i^2n^2-i^2n^2)/2

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

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

\frac{e ^{4} i ^{2} n ^{2} - i ^{2} n ^{2}}{2}

Solution steps

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

Apply Integration By Parts

: [i^{2} n^{2} x^{2} - \int i^{2} n^{2} x d x]_{1}^{e^{2}}

= [i^{2} n^{2} x^{2} - \int i^{2} n^{2} x d x]_{1}^{e^{2}}

\int i^{2} n^{2} x d x = \frac{i ^{2} n ^{2} x ^{2}}{2}

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

Join

i^{2} n^{2} x^{2} - \frac{i ^{2} n ^{2} x ^{2}}{2} : \frac{i ^{2} n ^{2} x ^{2}}{2}

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

Compute the boundaries:

\frac{e ^{4} i ^{2} n ^{2}}{2} - \frac{i ^{2} n ^{2}}{2}

= \frac{e ^{4} i ^{2} n ^{2}}{2} - \frac{i ^{2} n ^{2}}{2}

Simplify

= \frac{e ^{4} i ^{2} n ^{2} - 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^4i^2n^2-i^2n^2)/2