What is the integral of (ln(x))^2 1/x ?

The integral of (ln(x))^2 1/x is (ln^3(x))/3+C

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integral of (ln(x))^2 1/x

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

\frac{ln ^{3} ( x )}{3} + C

Solution steps

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

Simplify

(ln (x))^{2} \frac{1}{x} : \frac{ln ^{2} ( x )}{x}

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

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{u ^{3}}{3}

Substitute back

u = ln (x)

= \frac{ln ^{3} ( x )}{3}

Add a constant to the solution

= \frac{ln ^{3} ( x )}{3} + C

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What is the integral of (ln(x))^2 1/x ?
The integral of (ln(x))^2 1/x is (ln^3(x))/3+C