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

The integral of ((3x-2))/((2x-1)) is 3/2 x-3/4-1/4 ln|2x-1|+C

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integral of ((3x-2))/((2x-1))

integral

\frac{( 3 x - 2 )}{( 2 x - 1 )}

\frac{3}{2} x - \frac{3}{4} - \frac{1}{4} ln ∣ 2 x - 1 ∣ + C

Solution steps

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

Expand

\frac{3 x - 2}{2 x - 1} : \frac{3 x}{2 x - 1} - \frac{2}{2 x - 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{3 x}{2 x - 1} d x = \frac{3}{4} (2 x - 1 + ln ∣ 2 x - 1 ∣)

\int \frac{2}{2 x - 1} d x = ln ∣ 2 x - 1 ∣

= \frac{3}{4} (2 x - 1 + ln ∣ 2 x - 1 ∣) - ln ∣ 2 x - 1 ∣

Simplify

\frac{3}{4} (2 x - 1 + ln ∣ 2 x - 1 ∣) - ln ∣ 2 x - 1 ∣ : \frac{3}{2} x - \frac{3}{4} - \frac{1}{4} ln ∣ 2 x - 1 ∣

= \frac{3}{2} x - \frac{3}{4} - \frac{1}{4} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= \frac{3}{2} x - \frac{3}{4} - \frac{1}{4} ln ∣ 2 x - 1 ∣ + C

x - \frac{1}{2} cos (x^{\frac{1}{2}})

- 2 cos (2 x)

\frac{( 16 - x )}{( 4 + x )}

\frac{1}{2 + x ^{2}}

sin^{2} (x) - cos^{2} (x)

What is the integral of ((3x-2))/((2x-1)) ?
The integral of ((3x-2))/((2x-1)) is 3/2 x-3/4-1/4 ln|2x-1|+C