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

The integral of (3(2+ln(-3x))^3)/x is 3/4 (2+ln(-3x))^4+C

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

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

\frac{3}{4} (2 + ln (- 3 x))^{4} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{( 2 + ln ( u ) ) ^{3}}{u} d u

= 3 \cdot \int \frac{( 2 + ln ( u ) ) ^{3}}{u} d u

Apply u-substitution

: \int v^{3} d v

= 3 \cdot \int v^{3} d v

Apply the Power Rule

: \frac{v ^{4}}{4}

= 3 \cdot \frac{v ^{4}}{4}

Substitute back

= 3 \cdot \frac{( 2 + ln ( - 3 x ) ) ^{4}}{4}

Simplify

3 \cdot \frac{( 2 + ln ( - 3 x ) ) ^{4}}{4} : \frac{3}{4} (2 + ln (- 3 x))^{4}

= \frac{3}{4} (2 + ln (- 3 x))^{4}

Add a constant to the solution

= \frac{3}{4} (2 + ln (- 3 x))^{4} + C

\int 18 d x

x \to 0 lim (\frac{x ^{5} + 4 x}{x})

f (x) = \frac{1}{3} (x^{2} + 2)^{\frac{3}{2}}

x \to - 1.1 lim (\frac{1}{x ^{2}})

What is the integral of (3(2+ln(-3x))^3)/x ?
The integral of (3(2+ln(-3x))^3)/x is 3/4 (2+ln(-3x))^4+C