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

f (x) = \frac{x}{1 + x ^{2}}

\int_{0}^{1} (x x^{2} + 1) d x

\frac{d}{d x} (\frac{120}{( 1 - x ) ^{6}})

\frac{( x + 3 )}{2 x + 1}

(- 8.5) (6.2)

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