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

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

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

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

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

Solution steps

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

\frac{d}{d y} (\frac{ln ( y )}{2 y} + \frac{1}{y})

x \to - 2 lim (- 10)

x y^{^{'}} + y = y^{2}, y (1) = - 9

\frac{d}{d x} (cosh (3 x))

x \to 1 - lim (\frac{x + 1}{x - 1})

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