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

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

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

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

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

Solution steps

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

Simplify

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

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

(ln (x + 2))^{^{'}}

x \to 7 - lim (\frac{6 x}{x - 7})

\frac{d}{d x} (- 1.2 x^{2} + 9.6 x)

\frac{( 1 - e ^{- 5 s} )}{s ^{2}}

\int \frac{1}{x ln ( x ) ln ^{2} ( x ) - 4} d x

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