What is the integral of 1/(3-y) ?

The integral of 1/(3-y) is -ln|3-y|+C

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integral of 1/(3-y)

\int \frac{1}{3 - y} d y

- ln ∣ 3 - y ∣ + C

Solution steps

\int \frac{1}{3 - y} d y

Apply u-substitution

: \int - \frac{1}{u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = 3 - y

= - ln ∣ 3 - y ∣

Add a constant to the solution

= - ln ∣ 3 - y ∣ + C

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\frac{d y}{d x} = \frac{( x ^{2} + y ^{2} )}{x y}

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