What is the integral of x/(9x-8) ?

The integral of x/(9x-8) is 1/81 (9x-8+8ln|9x-8|)+C

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integral of x/(9x-8)

\int \frac{x}{9 x - 8} d x

\frac{1}{81} (9 x - 8 + 8 ln ∣ 9 x - 8 ∣) + C

Solution steps

\int \frac{x}{9 x - 8} d x

Apply u-substitution

: \int \frac{u + 8}{81 u} d u

= \int \frac{u + 8}{81 u} d u

Take the constant out:

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

= \frac{1}{81} \cdot \int \frac{u + 8}{u} d u

Expand

\frac{u + 8}{u} : 1 + \frac{8}{u}

= \frac{1}{81} \cdot \int 1 + \frac{8}{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}{81} (\int 1 d u + \int \frac{8}{u} d u)

\int 1 d u = u

\int \frac{8}{u} d u = 8 ln ∣ u ∣

= \frac{1}{81} (u + 8 ln ∣ u ∣)

Substitute back

u = 9 x - 8

= \frac{1}{81} (9 x - 8 + 8 ln ∣ 9 x - 8 ∣)

Add a constant to the solution

= \frac{1}{81} (9 x - 8 + 8 ln ∣ 9 x - 8 ∣) + C

sin^{3} (2 x)

\frac{d}{d x} (\frac{2 x ^{2} + 6 x + 8}{x})

\frac{d}{d x} ((x^{2} + 5 x + 2)^{7})

\frac{\partial}{\partial z} (x y e^{x})

What is the integral of x/(9x-8) ?
The integral of x/(9x-8) is 1/81 (9x-8+8ln|9x-8|)+C