What is the integral of (2x-4)/(x^3+4x^2) ?

The integral of (2x-4)/(x^3+4x^2) is 3/4 ln|x|-3/4 ln|x+4|+1/x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x-4)/(x^3+4x^2)

\int \frac{2 x - 4}{x ^{3} + 4 x ^{2}} d x

\frac{3}{4} ln ∣ x ∣ - \frac{3}{4} ln ∣ x + 4 ∣ + \frac{1}{x} + C

Solution steps

\int \frac{2 x - 4}{x ^{3} + 4 x ^{2}} d x

Expand

\frac{2 x - 4}{x ^{3} + 4 x ^{2}} : \frac{2 x}{x ^{3} + 4 x ^{2}} - \frac{4}{x ^{3} + 4 x ^{2}}

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{2 x}{x ^{3} + 4 x ^{2}} d x - \int \frac{4}{x ^{3} + 4 x ^{2}} d x

\int \frac{2 x}{x ^{3} + 4 x ^{2}} d x = 2 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣)

\int \frac{4}{x ^{3} + 4 x ^{2}} d x = - \frac{1}{4} ln ∣ x ∣ - \frac{1}{x} + \frac{1}{4} ln ∣ x + 4 ∣

= 2 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣) - (- \frac{1}{4} ln ∣ x ∣ - \frac{1}{x} + \frac{1}{4} ln ∣ x + 4 ∣)

Simplify

2 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣) - (- \frac{1}{4} ln ∣ x ∣ - \frac{1}{x} + \frac{1}{4} ln ∣ x + 4 ∣) : \frac{3}{4} ln ∣ x ∣ - \frac{3}{4} ln ∣ x + 4 ∣ + \frac{1}{x}

= \frac{3}{4} ln ∣ x ∣ - \frac{3}{4} ln ∣ x + 4 ∣ + \frac{1}{x}

Add a constant to the solution

= \frac{3}{4} ln ∣ x ∣ - \frac{3}{4} ln ∣ x + 4 ∣ + \frac{1}{x} + C

\frac{d}{d x} (3 x^{- \frac{5}{3}})

\frac{d}{d x} (cos (sin (2 x)))

\frac{d}{d x} (a cos (7 x) + b sin (7 x))

n = 1 \sum \infty \frac{2 ^{- n}}{5 ^{- n}}

\frac{d}{d x} (5 x + 3)

What is the integral of (2x-4)/(x^3+4x^2) ?
The integral of (2x-4)/(x^3+4x^2) is 3/4 ln|x|-3/4 ln|x+4|+1/x+C