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

The integral from 1 to e^3 of (ln^2(x^2))/x is 36

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

\int_{1}^{e^{3}} \frac{ln ^{2} ( x ^{2} )}{x} d x

36

Solution steps

\int_{1}^{e^{3}} \frac{ln ^{2} ( x ^{2} )}{x} d x

Apply u-substitution

= \int_{1}^{e^{6}} \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^{6}} \frac{ln ^{2} ( u )}{u} d u

Apply u-substitution

= \frac{1}{2} \cdot \int_{0}^{6} v^{2} d v

Apply the Power Rule

= \frac{1}{2} [\frac{v ^{3}}{3}]_{0}^{6}

Compute the boundaries:

72

= \frac{1}{2} \cdot 72

Simplify

= 36

What is the integral from 1 to e^3 of (ln^2(x^2))/x ?
The integral from 1 to e^3 of (ln^2(x^2))/x is 36