What is the integral of (3(-5+ln(4x))^3)/x ?

The integral of (3(-5+ln(4x))^3)/x is 3(-125ln|x|+75/2 ln^2(4x)-5ln^3(4x)+(ln^4(4x))/4)+C

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integral of (3(-5+ln(4x))^3)/x

\int \frac{3 ( - 5 + ln ( 4 x ) ) ^{3}}{x} d x

3 (- 125 ln ∣ x ∣ + \frac{75}{2} ln^{2} (4 x) - 5 ln^{3} (4 x) + \frac{ln ^{4} ( 4 x )}{4}) + C

Solution steps

\int \frac{3 ( - 5 + ln ( 4 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{( - 5 + ln ( 4 x ) ) ^{3}}{x} d x

Expand

\frac{( - 5 + ln ( 4 x ) ) ^{3}}{x} : - \frac{125}{x} + \frac{75 ln ( 4 x )}{x} - \frac{15 ln ^{2} ( 4 x )}{x} + \frac{ln ^{3} ( 4 x )}{x}

= 3 \cdot \int - \frac{125}{x} + \frac{75 ln ( 4 x )}{x} - \frac{15 ln ^{2} ( 4 x )}{x} + \frac{ln ^{3} ( 4 x )}{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (- \int \frac{125}{x} d x + \int \frac{75 ln ( 4 x )}{x} d x - \int \frac{15 ln ^{2} ( 4 x )}{x} d x + \int \frac{ln ^{3} ( 4 x )}{x} d x)

\int \frac{125}{x} d x = 125 ln ∣ x ∣

\int \frac{75 ln ( 4 x )}{x} d x = \frac{75}{2} ln^{2} (4 x)

\int \frac{15 ln ^{2} ( 4 x )}{x} d x = 5 ln^{3} (4 x)

\int \frac{ln ^{3} ( 4 x )}{x} d x = \frac{ln ^{4} ( 4 x )}{4}

= 3 (- 125 ln ∣ x ∣ + \frac{75}{2} ln^{2} (4 x) - 5 ln^{3} (4 x) + \frac{ln ^{4} ( 4 x )}{4})

Add a constant to the solution

= 3 (- 125 ln ∣ x ∣ + \frac{75}{2} ln^{2} (4 x) - 5 ln^{3} (4 x) + \frac{ln ^{4} ( 4 x )}{4}) + C

f (x) = x^{2} - 9 x + 18, 4, at x = 3

\frac{d}{d x} (\frac{x}{( 3 x - 8 ) ^{8}})

\frac{4 s}{s ^{2} + 16}

f (x) = \frac{3}{x + 3}, at x = - \frac{1}{3}

y = x^{2}, y = 8 - 2 x

What is the integral of (3(-5+ln(4x))^3)/x ?
The integral of (3(-5+ln(4x))^3)/x is 3(-125ln|x|+75/2 ln^2(4x)-5ln^3(4x)+(ln^4(4x))/4)+C