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

The integral from 1 to e of ((ln(x^2))^2)/x is 4/3

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

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

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

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

Simplify

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

Apply u-substitution

: \int_{1}^{e^{2}} \frac{ln ^{2} ( u )}{2 u} d u

= \int_{1}^{e^{2}} \frac{ln ^{2} ( u )}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{e^{2}} \frac{ln ^{2} ( u )}{u} d u

Apply u-substitution

: \int_{0}^{2} v^{2} d v

= \frac{1}{2} \cdot \int_{0}^{2} v^{2} d v

Apply the Power Rule

: [\frac{v ^{3}}{3}]_{0}^{2}

= \frac{1}{2} [\frac{v ^{3}}{3}]_{0}^{2}

Compute the boundaries:

\frac{8}{3}

= \frac{1}{2} \cdot \frac{8}{3}

Simplify

= \frac{4}{3}

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What is the integral from 1 to e of ((ln(x^2))^2)/x ?
The integral from 1 to e of ((ln(x^2))^2)/x is 4/3