What is the integral of 1/(4-y) ?

The integral of 1/(4-y) is -ln|4-y|+C

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integral of 1/(4-y)

\int \frac{1}{4 - y} d y

- ln ∣ 4 - y ∣ + C

Solution steps

\int \frac{1}{4 - 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 = 4 - y

= - ln ∣ 4 - y ∣

Add a constant to the solution

= - ln ∣ 4 - y ∣ + C

x \to 8 + lim (\frac{x - 8}{∣ x - 8 ∣})

y = 3 x^{2} - 2 x, at x = - 1

\frac{d}{d x} (x^{2} (x - 2)^{4})

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

x \to \infty lim (\frac{3}{3})