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

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

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

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

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

Solution steps

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

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