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

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

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

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

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

Solution steps

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

Simplify

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

Add a constant to the solution

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

x \to a lim ([2 f (x) - 5 g (x)])

\int 4 sec (x) d x

y = 3 - 2 x^{2}

2 x + 1

\frac{d}{d x} (e^{3 x} sin (x^{2}))

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