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

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

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integral of (-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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \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})

Simplify

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

Add a constant to the solution

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

3 + y^{2}

y^{^{''}} - 2 y^{^{'}} = (4 t + 9) sin (2 t)

\int 2 sec^{2} (x) d x

\int_{- 1}^{0} \frac{1}{x} d x

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

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