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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (9 x^{5})

\int (ln (17 x))^{2} d x

x \to \infty lim (\frac{8}{10})

f (x) = \frac{x ^{2} + 5}{x - 2}

\frac{d y}{d x} = \frac{1}{x}

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