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

The integral from 1 to e of ((2+ln(x))^2)/x is 19/3

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

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

\frac{19}{3}

Decimal Notation

6.33333 \dots

Solution steps

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

Apply Integration By Parts

: [(2 + ln (x))^{2} - \int \frac{- ln ^{2} ( x ) - 2 ln ( x )}{x} d x]_{1}^{e}

= [(2 + ln (x))^{2} - \int \frac{- ln ^{2} ( x ) - 2 ln ( x )}{x} d x]_{1}^{e}

\int \frac{- ln ^{2} ( x ) - 2 ln ( x )}{x} d x = - \frac{ln ^{3} ( x )}{3} - ln^{2} (x)

= [(2 + ln (x))^{2} - (- \frac{ln ^{3} ( x )}{3} - ln^{2} (x))]_{1}^{e}

Simplify

[(2 + ln (x))^{2} - (- \frac{ln ^{3} ( x )}{3} - ln^{2} (x))]_{1}^{e} : [\frac{1}{3} (12 ln (x) + 6 ln^{2} (x) + ln^{3} (x) + 12)]_{1}^{e}

= [\frac{1}{3} (12 ln (x) + 6 ln^{2} (x) + ln^{3} (x) + 12)]_{1}^{e}

Compute the boundaries:

\frac{19}{3}

= \frac{19}{3}

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What is the integral from 1 to e of ((2+ln(x))^2)/x ?
The integral from 1 to e of ((2+ln(x))^2)/x is 19/3