What is the integral of 1/(3-2y) ?

The integral of 1/(3-2y) is -1/2 ln|3-2y|+C

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integral of 1/(3-2y)

\int \frac{1}{3 - 2 y} d y

- \frac{1}{2} ln ∣ 3 - 2 y ∣ + C

Solution steps

\int \frac{1}{3 - 2 y} d y

Apply u-substitution

: \int - \frac{1}{2 u} d u

= \int - \frac{1}{2 u} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = 3 - 2 y

= - \frac{1}{2} ln ∣ 3 - 2 y ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ 3 - 2 y ∣ + C

\frac{\partial}{\partial x} ((x)^{x})

3 e^{- x}

\frac{\partial}{\partial y} (sin (\frac{x}{1 + y}))

n = 0 \sum \infty \frac{1}{n ^{6}}

\int \frac{x}{( 1 - x )} d x