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

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

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integral of 1/(3-2x)

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

- \frac{1}{2} ln ∣ 3 - 2 x ∣ + C

Solution steps

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

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 x

= - \frac{1}{2} ln ∣ 3 - 2 x ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ 3 - 2 x ∣ + C

\int \frac{1}{1 + 4 x ^{2}} d x

\frac{d y}{d x} + x y^{2} = 0

\int_{0}^{4} \frac{1}{x ^{2} - 6 x + 5} d x

(e^{- 3 x})^{^{'}}

\frac{d}{d x} (ln (\frac{x + 1}{x ^{3} + 1}))