What is the integral of ((ln(x)-1)^2)/x ?

The integral of ((ln(x)-1)^2)/x is (ln^3(x))/3-ln^2(x)+ln|x|+C

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integral of ((ln(x)-1)^2)/x

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

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

Solution steps

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

Expand

\frac{( ln ( x ) - 1 ) ^{2}}{x} : \frac{ln ^{2} ( x )}{x} - \frac{2 ln ( x )}{x} + \frac{1}{x}

= \int \frac{ln ^{2} ( x )}{x} - \frac{2 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{ln ^{2} ( x )}{x} d x - \int \frac{2 ln ( x )}{x} d x + \int \frac{1}{x} d x

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

Add a constant to the solution

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

y = \frac{x}{3} + \frac{3 x ^{2}}{2}

x \to 0 lim (\frac{( 1 )}{x ^{2}})

\int \frac{( x ^{2} - x + 12 )}{x ^{3} + 6 x} d x

(1 + cos (x)) \frac{d y}{d x} = (1 + e^{- y}) sin (x)

What is the integral of ((ln(x)-1)^2)/x ?
The integral of ((ln(x)-1)^2)/x is (ln^3(x))/3-ln^2(x)+ln|x|+C