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

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

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

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

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

Solution steps

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

Simplify

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

x \to π lim (6)

\frac{d ^{2} Q}{d t ^{2}} - 7 \frac{d Q}{d t} + 5 Q = 0

\int \frac{cos ( t )}{1 + sin ^{2} ( t )} d t

\frac{d}{d x} (\frac{7}{( x + 1 ) ^{2}})

\frac{d}{d x} (\frac{1}{1 + \frac{1}{4} x})

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