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

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

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integral of 1/x ln^2(x)

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

\frac{x ln ^{2} ( x ) - 2 ( x ln ( x ) - x )}{x} + \frac{1}{3} ln^{3} (x) - ln^{2} (x) + 2 ln ∣ x ∣ + C

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{x ln ^{2} ( x ) - 2 ( x ln ( x ) - x )}{x ^{2}} d x = - \frac{1}{3} ln^{3} (x) + ln^{2} (x) - 2 ln ∣ x ∣

= \frac{x ln ^{2} ( x ) - 2 ( x ln ( x ) - x )}{x} - (- \frac{1}{3} ln^{3} (x) + ln^{2} (x) - 2 ln ∣ x ∣)

Simplify

= \frac{x ln ^{2} ( x ) - 2 ( x ln ( x ) - x )}{x} + \frac{1}{3} ln^{3} (x) - ln^{2} (x) + 2 ln ∣ x ∣

Add a constant to the solution

= \frac{x ln ^{2} ( x ) - 2 ( x ln ( x ) - x )}{x} + \frac{1}{3} ln^{3} (x) - ln^{2} (x) + 2 ln ∣ x ∣ + C

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