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

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

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

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

\frac{3}{2}

Decimal

1.5

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

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

Compute the boundaries:

\frac{3}{2}

= \frac{3}{2}

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