What is the integral of 1/(b-ay) ?

The integral of 1/(b-ay) is -1/a ln|b-ay|+C

integral of 1/(b-ay)

\int \frac{1}{b - a y} d y

- \frac{1}{a} ln ∣ b - a y ∣ + C

Solution steps

\int \frac{1}{b - a y} d y

Apply u-substitution

= \int - \frac{1}{a u} d u

Take the constant out:

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

= - \frac{1}{a} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{a} ln ∣ u ∣

Substitute back

u = b - a y

= - \frac{1}{a} ln ∣ b - a y ∣

Add a constant to the solution

= - \frac{1}{a} ln ∣ b - a y ∣ + C

\int (x^{2} - 4 x + 5) d x

\frac{d}{d x} (arcsin (\frac{x}{10}))

\frac{\partial}{\partial x} (- \frac{4}{x ^{2} y})

d er i v a t i v e f (t) = t^{\frac{1}{2}} (13 t + 2)

d er i v a t i v e \frac{9}{( 9 - x ^{8} ) ^{\frac{3}{2}}}