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

The integral from 1 to e of ln(2x) is eln(2e)-e-ln(2)+1

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integral from 1 to e of ln(2x)

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

e ln (2 e) - e - ln (2) + 1

Decimal Notation

2.19102 \dots

Solution steps

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

Apply Integration By Parts

: [x ln (2 x) - \int 1 d x]_{1}^{e}

= [x ln (2 x) - \int 1 d x]_{1}^{e}

\int 1 d x = x

= [x ln (2 x) - x]_{1}^{e}

Compute the boundaries:

e ln (2 e) - e - ln (2) + 1

= e ln (2 e) - e - ln (2) + 1

6 y^{^{''}} + y^{^{'}} = 0

\frac{d}{d x} (\frac{π}{2})

\int 5 e^{- 4 x + 8} - 5 d x

\int y e^{- 2 y} d y

\frac{d}{d t} (\frac{1}{2} t^{2} + t)

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