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

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

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

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

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

Decimal

1.19314 \dots

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

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

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

\frac{d}{d t} (cos^{3} (3 t^{2}))

\frac{v}{v - 1}

x \to 0 lim (\frac{sin ( x )}{5 x})

\int (x^{3} + 1)^{2} \cdot x^{2} d x

\frac{d y}{d x} + 3 y = x

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