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 (1 - x) (1 - x^{2}) d x

\int (x - 1)^{2} (x - 2) d x

\int \frac{x ^{3} + 1}{x x} d x

\int x^{2} e^{\frac{x ^{2}}{2}} d x

\int cos (2 - x) d x

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