What is the integral of 1/(x^2)ln(3x) ?

The integral of 1/(x^2)ln(3x) is -(ln(3x))/x-1/x+C

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integral of 1/(x^2)ln(3x)

\int \frac{1}{x ^{2}} ln (3 x) d x

- \frac{ln ( 3 x )}{x} - \frac{1}{x} + C

Solution steps

\int \frac{1}{x ^{2}} ln (3 x) d x

Simplify

\frac{1}{x ^{2}} ln (3 x) : \frac{ln ( 3 x )}{x ^{2}}

= \int \frac{ln ( 3 x )}{x ^{2}} d x

Apply Integration By Parts

: - \frac{ln ( 3 x )}{x} - \int - \frac{1}{x ^{2}} d x

= - \frac{ln ( 3 x )}{x} - \int - \frac{1}{x ^{2}} d x

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

= - \frac{ln ( 3 x )}{x} - \frac{1}{x}

Add a constant to the solution

= - \frac{ln ( 3 x )}{x} - \frac{1}{x} + C

\frac{d y}{d x}, y = x e^{y}

x \to 6 - lim (sec (\frac{π x}{4}))

\frac{\partial}{\partial y} (2 x - 3 y)

\frac{\partial}{\partial y} (\frac{1}{x y})

y^{^{'}} = \frac{15 x}{y}

What is the integral of 1/(x^2)ln(3x) ?
The integral of 1/(x^2)ln(3x) is -(ln(3x))/x-1/x+C