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

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

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

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

e - 2

Decimal Notation

0.71828 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int - 1 d x = - x

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

Simplify

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

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

Compute the boundaries:

e - 2

= e - 2

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\int_{0}^{16} 4 x d x

\int_{s i n (x)}^{c o s (x)} 1 d y

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