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

The integral of x/(x^3+x^2+4x+4) is -1/5 ln|x+1|+1/5 (1/2 ln|(x^2)/4+1|+2arctan(x/2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Take the partial fraction of

\frac{x}{x ^{3} + x ^{2} + 4 x + 4} : - \frac{1}{5 ( x + 1 )} + \frac{x + 4}{5 ( x ^{2} + 4 )}

= \int - \frac{1}{5 ( x + 1 )} + \frac{x + 4}{5 ( x ^{2} + 4 )} d 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{1}{5 ( x + 1 )} d x + \int \frac{x + 4}{5 ( x ^{2} + 4 )} d x

\int \frac{1}{5 ( x + 1 )} d x = \frac{1}{5} ln ∣ x + 1 ∣

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

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

Add a constant to the solution

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

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

\int (tan (x) + sec (x))^{2} d x

\int - csc (2 x) cot (2 x) d x

h \to 0 lim (\frac{( 5 + h ) ^{4} - 625}{h})

d er i v a t i v e 3 e^{3 t}

What is the integral of x/(x^3+x^2+4x+4) ?
The integral of x/(x^3+x^2+4x+4) is -1/5 ln|x+1|+1/5 (1/2 ln|(x^2)/4+1|+2arctan(x/2))+C