What is the integral from 0 to e of (ln(x^3))/x ?

The integral from 0 to e of (ln(x^3))/x is diverges

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integral from 0 to e of (ln(x^3))/x

\int_{0}^{e} \frac{ln ( x ^{3} )}{x} d x

diverges

Solution steps

\int_{0}^{e} \frac{ln ( x ^{3} )}{x} d x

Apply u-substitution

: \int_{0}^{e^{3}} \frac{ln ( u )}{3 u} d u

= \int_{0}^{e^{3}} \frac{ln ( u )}{3 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{0}^{e^{3}} \frac{ln ( u )}{u} d u

Apply Integration By Parts

: [ln (u) - \int \frac{1 - ln ( u )}{u} d u]_{0}^{e^{3}}

= \frac{1}{3} [ln (u) - \int \frac{1 - ln ( u )}{u} d u]_{0}^{e^{3}}

\int \frac{1 - ln ( u )}{u} d u = - \frac{( 1 - ln ( u ) ) ^{2}}{2}

= \frac{1}{3} [ln (u) - (- \frac{( 1 - ln ( u ) ) ^{2}}{2})]_{0}^{e^{3}}

Simplify

\frac{1}{3} [ln (u) - (- \frac{( 1 - ln ( u ) ) ^{2}}{2})]_{0}^{e^{3}} : \frac{1}{3} [\frac{ln ^{2} ( u ) + 1}{2}]_{0}^{e^{3}}

= \frac{1}{3} [\frac{ln ^{2} ( u ) + 1}{2}]_{0}^{e^{3}}

Compute the boundaries:

diverges

= diverges

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What is the integral from 0 to e of (ln(x^3))/x ?
The integral from 0 to e of (ln(x^3))/x is diverges