What is the integral of (5x^2-4x+4)/(x^3+4x) ?

The integral of (5x^2-4x+4)/(x^3+4x) is 2ln|x^2+4|-2arctan(x/2)+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 (5x^2-4x+4)/(x^3+4x)

\int \frac{5 x ^{2} - 4 x + 4}{x ^{3} + 4 x} d x

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

Solution steps

\int \frac{5 x ^{2} - 4 x + 4}{x ^{3} + 4 x} d x

Expand

\frac{5 x ^{2} - 4 x + 4}{x ^{3} + 4 x} : \frac{5 x ^{2}}{x ^{3} + 4 x} - \frac{4 x}{x ^{3} + 4 x} + \frac{4}{x ^{3} + 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{5 x ^{2}}{x ^{3} + 4 x} d x - \int \frac{4 x}{x ^{3} + 4 x} d x + \int \frac{4}{x ^{3} + 4 x} d x

\int \frac{5 x ^{2}}{x ^{3} + 4 x} d x = \frac{5}{2} ln x^{2} + 4

\int \frac{4 x}{x ^{3} + 4 x} d x = 2 arctan (\frac{x}{2})

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

= \frac{5}{2} ln x^{2} + 4 - 2 arctan (\frac{x}{2}) + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4

Simplify

\frac{5}{2} ln x^{2} + 4 - 2 arctan (\frac{x}{2}) + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4 : 2 ln x^{2} + 4 - 2 arctan (\frac{x}{2}) + ln ∣ x ∣

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

Add a constant to the solution

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

\int_{2}^{7} t^{2} ln (3 t) d t

\int - e^{- 2 c s c (2 x)} cot (2 x) csc (2 x) d x

\int \frac{1}{( x - x ^{2} )} d x

y^{^{'}} + \frac{x}{x ^{2} + 1} y = - 4 x

y = (\frac{( x + 2 )}{( x - 3 )})^{2}, at x = 4

What is the integral of (5x^2-4x+4)/(x^3+4x) ?
The integral of (5x^2-4x+4)/(x^3+4x) is 2ln|x^2+4|-2arctan(x/2)+ln|x|+C