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

The integral from 1 to e of ln(x)-ln^2(x) is -e+3

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

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

- e + 3

Decimal Notation

0.28171 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int 1 - 2 ln (x) d x = 3 x - 2 x ln (x)

= [x (ln (x) - ln^{2} (x)) - (3 x - 2 x ln (x))]_{1}^{e}

Expand

x (ln (x) - ln^{2} (x)) - (3 x - 2 x ln (x)) : 3 x ln (x) - x ln^{2} (x) - 3 x

= [3 x ln (x) - x ln^{2} (x) - 3 x]_{1}^{e}

Compute the boundaries:

- e + 3

= - e + 3

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What is the integral from 1 to e of ln(x)-ln^2(x) ?
The integral from 1 to e of ln(x)-ln^2(x) is -e+3