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

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

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integral from 1 to 2 of (1+ln(x))/x

\int_{1}^{2} \frac{1 + ln ( x )}{x} d x

\frac{ln ^{2} ( 2 ) + 2 ln ( 2 )}{2}

Solution steps

\int_{1}^{2} \frac{1 + ln ( x )}{x} d x

Apply u-substitution

: \int_{1}^{1 + l n (2)} u d u

= \int_{1}^{1 + l n (2)} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{1}^{1 + l n (2)}

= [\frac{u ^{2}}{2}]_{1}^{1 + l n (2)}

Compute the boundaries:

\frac{( 1 + ln ( 2 ) ) ^{2}}{2} - \frac{1}{2}

= \frac{( 1 + ln ( 2 ) ) ^{2}}{2} - \frac{1}{2}

Simplify

= \frac{ln ^{2} ( 2 ) + 2 ln ( 2 )}{2}

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