What is the integral of (4x)/(x-1) ?

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

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integral of (4x)/(x-1)

\int \frac{4 x}{x - 1} d x

4 (x - 1 + ln ∣ x - 1 ∣) + C

Solution steps

\int \frac{4 x}{x - 1} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{x - 1} d x

Apply u-substitution

: \int \frac{u + 1}{u} d u

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

Expand

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

= 4 \cdot \int 1 + \frac{1}{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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 4 (u + ln ∣ u ∣)

Substitute back

u = x - 1

= 4 (x - 1 + ln ∣ x - 1 ∣)

Add a constant to the solution

= 4 (x - 1 + ln ∣ x - 1 ∣) + C

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