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

The integral of (x^2-2x-4)/(x^2-4x) is (x^2)/(x-4)+ln|x-4|-(16)/(x-4)+ln|x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{x ^{2}}{x - 4} + ln ∣ x - 4 ∣ - \frac{16}{x - 4} + ln ∣ x ∣ + C

Solution steps

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

Expand

\frac{x ^{2} - 2 x - 4}{x ^{2} - 4 x} : \frac{x ^{2}}{x ^{2} - 4 x} - \frac{2 x}{x ^{2} - 4 x} - \frac{4}{x ^{2} - 4 x}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{x ^{2}}{x ^{2} - 4 x} d x = \frac{x ^{2}}{x - 4} + 4 (ln ∣ x - 4 ∣ - \frac{4}{x - 4})

\int \frac{2 x}{x ^{2} - 4 x} d x = 2 ln ∣ x - 4 ∣

\int \frac{4}{x ^{2} - 4 x} d x = - ln ∣ x ∣ + ln ∣ x - 4 ∣

= \frac{x ^{2}}{x - 4} + 4 (ln ∣ x - 4 ∣ - \frac{4}{x - 4}) - 2 ln ∣ x - 4 ∣ - (- ln ∣ x ∣ + ln ∣ x - 4 ∣)

Simplify

\frac{x ^{2}}{x - 4} + 4 (ln ∣ x - 4 ∣ - \frac{4}{x - 4}) - 2 ln ∣ x - 4 ∣ - (- ln ∣ x ∣ + ln ∣ x - 4 ∣) : \frac{x ^{2}}{x - 4} + ln ∣ x - 4 ∣ - \frac{16}{x - 4} + ln ∣ x ∣

= \frac{x ^{2}}{x - 4} + ln ∣ x - 4 ∣ - \frac{16}{x - 4} + ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{x - 4} + ln ∣ x - 4 ∣ - \frac{16}{x - 4} + ln ∣ x ∣ + C

\int_{- 1}^{1} (e^{x} - (x^{2} - 1)) d x

\int \frac{3 x + 1}{x ^{2} + 1} d x

\frac{\partial}{\partial y} (x y e^{- 1 y})

x \to 1 lim (sin (\frac{π x}{2}))

\int - 3 \cdot x + 2 x d x

What is the integral of (x^2-2x-4)/(x^2-4x) ?
The integral of (x^2-2x-4)/(x^2-4x) is (x^2)/(x-4)+ln|x-4|-(16)/(x-4)+ln|x|+C