What is the integral from 0 to 1/(xy) of ln(y) ?

The integral from 0 to 1/(xy) of ln(y) is (-1-ln(xy))/(xy)

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integral from 0 to 1/(xy) of ln(y)

\int_{0}^{\frac{1}{x y}} ln (y) d y

\frac{- 1 - ln ( x y )}{x y}

Solution steps

\int_{0}^{\frac{1}{x y}} ln (y) d y

Apply Integration By Parts

= [y ln (y) - \int 1 d y]_{0}^{\frac{1}{x y}}

\int 1 d y = y

= [y ln (y) - y]_{0}^{\frac{1}{x y}}

Compute the boundaries:

- \frac{1}{x y} - \frac{ln ( x y )}{x y}

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

Simplify

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

\int_{0}^{2} (x + 2 x^{2} - x^{3}) d x

\int_{- 1}^{1} x^{2} - 3 x d x

\int_{0}^{2} e^{x^{2} - 2 x} d x

\int_{0}^{1} (x)^{2} d x

\int_{1}^{3} 2 e^{- 2 x} d x

What is the integral from 0 to 1/(xy) of ln(y) ?
The integral from 0 to 1/(xy) of ln(y) is (-1-ln(xy))/(xy)