What is the integral of (9x^2-8x+4)/(x^3+4x) ?

The integral of (9x^2-8x+4)/(x^3+4x) is 4ln|x^2+4|-4arctan(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 (9x^2-8x+4)/(x^3+4x)

\int \frac{9 x ^{2} - 8 x + 4}{x ^{3} + 4 x} d x

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

Solution steps

\int \frac{9 x ^{2} - 8 x + 4}{x ^{3} + 4 x} d x

Expand

\frac{9 x ^{2} - 8 x + 4}{x ^{3} + 4 x} : \frac{9 x ^{2}}{x ^{3} + 4 x} - \frac{8 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{9 x ^{2}}{x ^{3} + 4 x} d x - \int \frac{8 x}{x ^{3} + 4 x} d x + \int \frac{4}{x ^{3} + 4 x} d x

\int \frac{9 x ^{2}}{x ^{3} + 4 x} d x = \frac{9}{2} ln x^{2} + 4

\int \frac{8 x}{x ^{3} + 4 x} d x = 4 arctan (\frac{x}{2})

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

= \frac{9}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4

Simplify

\frac{9}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4 : 4 ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + ln ∣ x ∣

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (6 x^{4} y - 3 x y + 2 y)

x \to 2 lim (3 - x)

\int_{0}^{π} 2 π sin (x) d x

x y^{'} = 2 y

\int \frac{1}{36 - x ^{2}} d x

What is the integral of (9x^2-8x+4)/(x^3+4x) ?
The integral of (9x^2-8x+4)/(x^3+4x) is 4ln|x^2+4|-4arctan(x/2)+ln|x|+C