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

The integral from 1 to e of (ln(x)+1)/x is 3/2

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

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

\frac{3}{2}

Decimal Notation

1.5

Solution steps

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

Apply u-substitution

: \int_{1}^{2} u d u

= \int_{1}^{2} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{1}^{2}

= [\frac{u ^{2}}{2}]_{1}^{2}

Compute the boundaries:

\frac{3}{2}

= \frac{3}{2}

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

What is the integral from 1 to e of (ln(x)+1)/x ?
The integral from 1 to e of (ln(x)+1)/x is 3/2