What is the integral of x/(4x-3) ?

The integral of x/(4x-3) is 1/16 (4x-3+3ln|4x-3|)+C

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integral of x/(4x-3)

\int \frac{x}{4 x - 3} d x

\frac{1}{16} (4 x - 3 + 3 ln ∣ 4 x - 3 ∣) + C

Solution steps

\int \frac{x}{4 x - 3} d x

Apply u-substitution

: \int \frac{u + 3}{16 u} d u

= \int \frac{u + 3}{16 u} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{u + 3}{u} d u

Expand

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

= \frac{1}{16} \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}{16} (\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}{16} (u + 3 ln ∣ u ∣)

Substitute back

u = 4 x - 3

= \frac{1}{16} (4 x - 3 + 3 ln ∣ 4 x - 3 ∣)

Add a constant to the solution

= \frac{1}{16} (4 x - 3 + 3 ln ∣ 4 x - 3 ∣) + C

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What is the integral of x/(4x-3) ?
The integral of x/(4x-3) is 1/16 (4x-3+3ln|4x-3|)+C