What is the integral of (ln(y))/(xy) ?

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

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integral of (ln(y))/(xy)

\int \frac{ln ( y )}{x y} d y

\frac{1}{2 x} ln^{2} (y) + C

Solution steps

\int \frac{ln ( y )}{x y} d y

Take the constant out:

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

= \frac{1}{x} \cdot \int \frac{ln ( y )}{y} d y

Apply u-substitution

: \int u d u

= \frac{1}{x} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{1}{x} \cdot \frac{u ^{2}}{2}

Substitute back

u = ln (y)

= \frac{1}{x} \cdot \frac{ln ^{2} ( y )}{2}

Simplify

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

= \frac{1}{2 x} ln^{2} (y)

Add a constant to the solution

= \frac{1}{2 x} ln^{2} (y) + C

n = 1 \sum \infty \frac{1}{2 n ^{3}}

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

f (x) = x^{4} + 3 x

\frac{1}{1 - 2 x}

(sin (x) + x cos (x))^{^{'}}

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