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

The integral of x/(3x-4) is 1/9 (3x-4+4ln|3x-4|)+C

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integral of x/(3x-4)

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

\frac{1}{9} (3 x - 4 + 4 ln ∣ 3 x - 4 ∣) + C

Solution steps

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

Apply u-substitution

: \int \frac{u + 4}{9 u} d u

= \int \frac{u + 4}{9 u} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{u + 4}{u} d u

Expand

\frac{u + 4}{u} : 1 + \frac{4}{u}

= \frac{1}{9} \cdot \int 1 + \frac{4}{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}{9} (\int 1 d u + \int \frac{4}{u} d u)

\int 1 d u = u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= \frac{1}{9} (u + 4 ln ∣ u ∣)

Substitute back

u = 3 x - 4

= \frac{1}{9} (3 x - 4 + 4 ln ∣ 3 x - 4 ∣)

Add a constant to the solution

= \frac{1}{9} (3 x - 4 + 4 ln ∣ 3 x - 4 ∣) + C

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f (x) = (5 x^{3} + 4) (3 x^{7} - 5)

\int \frac{x + 4}{x ^{2} + 7 x - 8} d x

2 y^{^{''}} - 3 y^{^{'}} + y = e^{- 3 x}

\int tan (b x) d x

What is the integral of x/(3x-4) ?
The integral of x/(3x-4) is 1/9 (3x-4+4ln|3x-4|)+C