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

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

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integral of 3/(2x-1)

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

\frac{3}{2} ln ∣ 2 x - 1 ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \cdot \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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x - 1

= 3 \cdot \frac{1}{2} ln ∣ 2 x - 1 ∣

Simplify

3 \cdot \frac{1}{2} ln ∣ 2 x - 1 ∣ : \frac{3}{2} ln ∣ 2 x - 1 ∣

= \frac{3}{2} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= \frac{3}{2} ln ∣ 2 x - 1 ∣ + C

y^{^{''}} + 4 y = 8 t^{2} - 20 t + 8

\frac{d}{d x} (4 x^{3} + \frac{1}{x ^{2}} - x + 3)

\frac{d}{d x} (\frac{5 x}{( x ^{3} - 6 ) ^{2}})

\frac{\partial}{\partial x} (\frac{3}{2 x ^{2}})

y^{2} y^{^{'}} = x^{2}