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

- 72 x

\int \frac{y ^{2}}{1 - y} d y

f (x) = 3 x - 2 x^{- \frac{1}{2}} + 1, at x = 1

10 y^{^{''}} + 1000 y = 20 cos (8 t), y (0) = 0, y^{^{'}} (0) = 0

y = 3 x - 2

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