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

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

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integral of x/(2x-3)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u + 3}{4 u} d u

= \int \frac{u + 3}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{u + 3}{u} d u

Expand

\frac{u + 3}{u} : 1 + \frac{3}{u}

= \frac{1}{4} \cdot \int 1 + \frac{3}{u} d u

Apply the Sum Rule:

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

= \frac{1}{4} (\int 1 d u + \int \frac{3}{u} d u)

\int 1 d u = u

\int \frac{3}{u} d u = 3 ln ∣ u ∣

= \frac{1}{4} (u + 3 ln ∣ u ∣)

Substitute back

u = 2 x - 3

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (2 x y^{3} - 1)

\frac{d y}{d t} = \frac{t}{y + 1}

\frac{d}{d t} (e^{t} (a cos (2 t) + cos (2 t)))

f (x) = - x + \frac{4}{x} + 1

f (x) = - 3 cos (x) - \frac{x ^{2}}{2} + 20

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