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

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

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

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

1

Solution steps

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

Apply Integration By Parts

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

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

\int 1 d x = x

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

Compute the boundaries:

1

= 1

\int_{- 5}^{- 3} \frac{x}{x ^{2} - 9} d x

\int_{y}^{\infty} e^{- x} d x

\int_{0}^{x} y d y

\int_{y}^{1} y ln (x) d x

\int_{0}^{2} \frac{1}{x ^{2} + 1} d x

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