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

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

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

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

- 2 ln (2) + 2

Decimal Notation

0.61370 \dots

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

\int 1 d u = u

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

Compute the boundaries:

- (2 ln (2) - 2)

= - (2 ln (2) - 2)

Simplify

= - 2 ln (2) + 2

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