What is the integral of (ln^2(3x))/x ?

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

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integral of (ln^2(3x))/x

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

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

Solution steps

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

Apply u-substitution

: \int \frac{ln ^{2} ( u )}{u} d u

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

Apply u-substitution

: \int v^{2} d v

= \int v^{2} d v

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{v ^{3}}{3}

Substitute back

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (s u m)

y = - 5 x - 10, (- 4, 10)

\int sin^{3} (11 x) cos^{9} (11 x) d x

\frac{d}{d x} (\frac{x ^{2} - 1}{2})

y = \frac{x}{3} - \frac{3}{x}

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