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

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

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integral from 1 to e of (ln(4x))/x

\int_{1}^{e} \frac{ln ( 4 x )}{x} d x

\frac{ln ^{2} ( 4 e )}{2} - 2 ln^{2} (2)

Decimal Notation

1.88629 \dots

Solution steps

\int_{1}^{e} \frac{ln ( 4 x )}{x} d x

Apply u-substitution

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

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

Apply u-substitution

: \int_{2 l n (2)}^{l n (4 e)} v d v

= \int_{2 l n (2)}^{l n (4 e)} v d v

Apply the Power Rule

: [\frac{v ^{2}}{2}]_{2 l n (2)}^{l n (4 e)}

= [\frac{v ^{2}}{2}]_{2 l n (2)}^{l n (4 e)}

Compute the boundaries:

\frac{ln ^{2} ( 4 e )}{2} - 2 ln^{2} (2)

= \frac{ln ^{2} ( 4 e )}{2} - 2 ln^{2} (2)

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