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

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

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

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

e^{2} ln (2) + e^{2} - ln (2) + 1

Decimal Notation

12.81761 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int 1 d x = x

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

Compute the boundaries:

e^{2} ln (2) + e^{2} - ln (2) + 1

= e^{2} ln (2) + e^{2} - ln (2) + 1

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