What is the integral of x/(5x-4) ?

The integral of x/(5x-4) is 1/25 (5x-4+4ln|5x-4|)+C

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integral of x/(5x-4)

\int \frac{x}{5 x - 4} d x

\frac{1}{25} (5 x - 4 + 4 ln ∣ 5 x - 4 ∣) + C

Solution steps

\int \frac{x}{5 x - 4} d x

Apply u-substitution

: \int \frac{u + 4}{25 u} d u

= \int \frac{u + 4}{25 u} d u

Take the constant out:

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

= \frac{1}{25} \cdot \int \frac{u + 4}{u} d u

Expand

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

= \frac{1}{25} \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}{25} (\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}{25} (u + 4 ln ∣ u ∣)

Substitute back

u = 5 x - 4

= \frac{1}{25} (5 x - 4 + 4 ln ∣ 5 x - 4 ∣)

Add a constant to the solution

= \frac{1}{25} (5 x - 4 + 4 ln ∣ 5 x - 4 ∣) + C

\int x (a^{2} - x^{2}) d x

f (x) = \frac{1}{2} x^{2}

\frac{\partial}{\partial y} (ln (x^{8} y))

\frac{\partial}{\partial x} (x^{\frac{1}{4}} \cdot y^{\frac{3}{4}})

x \to π - lim (4 x)

What is the integral of x/(5x-4) ?
The integral of x/(5x-4) is 1/25 (5x-4+4ln|5x-4|)+C