What is the integral of (2x)/(x-4) ?

The integral of (2x)/(x-4) is 2(x-4+4ln|x-4|)+C

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integral of (2x)/(x-4)

\int \frac{2 x}{x - 4} d x

2 (x - 4 + 4 ln ∣ x - 4 ∣) + C

Solution steps

\int \frac{2 x}{x - 4} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{x - 4} d x

Apply u-substitution

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

= 2 \cdot \int \frac{u + 4}{u} d u

Expand

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

= 2 \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

= 2 (\int 1 d u + \int \frac{4}{u} d u)

\int 1 d u = u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= 2 (u + 4 ln ∣ u ∣)

Substitute back

u = x - 4

= 2 (x - 4 + 4 ln ∣ x - 4 ∣)

Add a constant to the solution

= 2 (x - 4 + 4 ln ∣ x - 4 ∣) + C

(x^{2} + 5) d x + (y - 3) (x - 2) d y = 0

\int 12 tan^{3} (x) d x

\int_{0}^{1} \frac{42}{4 y - 1} d y

n = 2 \sum \infty n \cdot (\frac{3}{4})^{n}

y^{^{''}} = 6 x + 2, y (1) = - 1, y^{^{'}} (1) = 7

What is the integral of (2x)/(x-4) ?
The integral of (2x)/(x-4) is 2(x-4+4ln|x-4|)+C