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

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

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

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

1

Solution steps

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

Apply u-substitution

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

Apply Integration By Parts

= [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

\int_{0}^{1} \frac{x}{1 + x} d x

\int_{0}^{2} x^{2} - 1 d x

\int_{\frac{π}{6}}^{\frac{π}{2}} cos^{2} (u) d u

\int_{0}^{\frac{π}{4}} (sec (t))^{2} d t

\int_{\frac{π}{4}}^{\frac{π}{2}} cot (x) d x

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