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

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

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

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

x ln^{2} (x) - 2 (x ln (x) - x) + C

Solution steps

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

Simplify

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

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

Apply Integration By Parts

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

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

\int 2 ln (x) d x = 2 (x ln (x) - x)

= x ln^{2} (x) - 2 (x ln (x) - x)

Add a constant to the solution

= x ln^{2} (x) - 2 (x ln (x) - x) + C

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