What is the integral of ((3ln(x)-1)^4)/x ?

The integral of ((3ln(x)-1)^4)/x is 81/5 ln^5(x)-27ln^4(x)+18ln^3(x)-6ln^2(x)+ln|x|+C

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integral of ((3ln(x)-1)^4)/x

\int \frac{( 3 ln ( x ) - 1 ) ^{4}}{x} d x

\frac{81}{5} ln^{5} (x) - 27 ln^{4} (x) + 18 ln^{3} (x) - 6 ln^{2} (x) + ln ∣ x ∣ + C

Solution steps

\int \frac{( 3 ln ( x ) - 1 ) ^{4}}{x} d x

Expand

\frac{( 3 ln ( x ) - 1 ) ^{4}}{x} : \frac{81 ln ^{4} ( x )}{x} - \frac{108 ln ^{3} ( x )}{x} + \frac{54 ln ^{2} ( x )}{x} - \frac{12 ln ( x )}{x} + \frac{1}{x}

= \int \frac{81 ln ^{4} ( x )}{x} - \frac{108 ln ^{3} ( x )}{x} + \frac{54 ln ^{2} ( x )}{x} - \frac{12 ln ( x )}{x} + \frac{1}{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

= \int \frac{81 ln ^{4} ( x )}{x} d x - \int \frac{108 ln ^{3} ( x )}{x} d x + \int \frac{54 ln ^{2} ( x )}{x} d x - \int \frac{12 ln ( x )}{x} d x + \int \frac{1}{x} d x

\int \frac{81 ln ^{4} ( x )}{x} d x = \frac{81}{5} ln^{5} (x)

\int \frac{108 ln ^{3} ( x )}{x} d x = 27 ln^{4} (x)

\int \frac{54 ln ^{2} ( x )}{x} d x = 18 ln^{3} (x)

\int \frac{12 ln ( x )}{x} d x = 6 ln^{2} (x)

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{81}{5} ln^{5} (x) - 27 ln^{4} (x) + 18 ln^{3} (x) - 6 ln^{2} (x) + ln ∣ x ∣

Add a constant to the solution

= \frac{81}{5} ln^{5} (x) - 27 ln^{4} (x) + 18 ln^{3} (x) - 6 ln^{2} (x) + ln ∣ x ∣ + C

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What is the integral of ((3ln(x)-1)^4)/x ?
The integral of ((3ln(x)-1)^4)/x is 81/5 ln^5(x)-27ln^4(x)+18ln^3(x)-6ln^2(x)+ln|x|+C