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

The integral from 1 to e^2 of (ln^2(x^2))/x is 32/3

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

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

\frac{32}{3}

Decimal Notation

10.66666 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{e^{4}} \frac{ln ^{2} ( u )}{2 u} d u

= \int_{1}^{e^{4}} \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^{4}} \frac{ln ^{2} ( u )}{u} d u

Apply u-substitution

: \int_{0}^{4} v^{2} d v

= \frac{1}{2} \cdot \int_{0}^{4} v^{2} d v

Apply the Power Rule

: [\frac{v ^{3}}{3}]_{0}^{4}

= \frac{1}{2} [\frac{v ^{3}}{3}]_{0}^{4}

Compute the boundaries:

\frac{64}{3}

= \frac{1}{2} \cdot \frac{64}{3}

Simplify

= \frac{32}{3}

\int e^{9 x} cos (4 x) d x

\int \frac{1}{csc ( 2 x )} csc (2 x) cot (2 x) d x

\frac{d y}{d x} + 3 x^{2} y = sin (x) e^{- x^{3}}, y (0) = 1

\int - 9 x^{5} e^{x^{6}} d x

x \to 5 - lim (\frac{3 x}{2 x + 10})

What is the integral from 1 to e^2 of (ln^2(x^2))/x ?
The integral from 1 to e^2 of (ln^2(x^2))/x is 32/3