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

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

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integral from 0 to e-1 of ln(1+x)

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

1

Solution steps

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

Apply u-substitution

: \int_{1}^{e} ln (u) d u

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

Apply Integration By Parts

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

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

\int 1 d u = u

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

Compute the boundaries:

1

= 1

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\int_{0}^{4} \frac{1}{( x - 2 ) ^{\frac{2}{3}}} d x

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