What is the integral from 0 to 1 of ln|x+1| ?

The integral from 0 to 1 of ln|x+1| is 2ln(2)-1

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integral from 0 to 1 of ln|x+1|

\int_{0}^{1} ln ∣ x + 1 ∣ d x

2 ln (2) - 1

Decimal

0.38629 \dots

Solution steps

\int_{0}^{1} ln ∣ x + 1 ∣ d x

Eliminate Absolutes

= \int_{0}^{1} ln (x + 1) d x

Apply u-substitution

= \int_{1}^{2} ln (u) d u

Apply Integration By Parts

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

\int 1 d u = u

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

Compute the boundaries:

2 ln (2) - 1

= 2 ln (2) - 1

What is the integral from 0 to 1 of ln|x+1| ?
The integral from 0 to 1 of ln|x+1| is 2ln(2)-1