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

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

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

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

1

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

2

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

Simplify

= 1

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