What is the integral of (3x^3+9x^2+2x+6)/(x^4+x^2) ?

The integral of (3x^3+9x^2+2x+6)/(x^4+x^2) is 2ln|x|-6/x+1/2 ln|x^2+1|+3arctan(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3x^3+9x^2+2x+6)/(x^4+x^2)

\int \frac{3 x ^{3} + 9 x ^{2} + 2 x + 6}{x ^{4} + x ^{2}} d x

2 ln ∣ x ∣ - \frac{6}{x} + \frac{1}{2} ln x^{2} + 1 + 3 arctan (x) + C

Solution steps

\int \frac{3 x ^{3} + 9 x ^{2} + 2 x + 6}{x ^{4} + x ^{2}} d x

Take the partial fraction of

\frac{3 x ^{3} + 9 x ^{2} + 2 x + 6}{x ^{4} + x ^{2}} : \frac{2}{x} + \frac{6}{x ^{2}} + \frac{x + 3}{x ^{2} + 1}

= \int \frac{2}{x} + \frac{6}{x ^{2}} + \frac{x + 3}{x ^{2} + 1} 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{2}{x} d x + \int \frac{6}{x ^{2}} d x + \int \frac{x + 3}{x ^{2} + 1} d x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{6}{x ^{2}} d x = - \frac{6}{x}

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

= 2 ln ∣ x ∣ - \frac{6}{x} + \frac{1}{2} ln x^{2} + 1 + 3 arctan (x)

Add a constant to the solution

= 2 ln ∣ x ∣ - \frac{6}{x} + \frac{1}{2} ln x^{2} + 1 + 3 arctan (x) + C

x \cdot sec (x)

\int \frac{1}{x 16 x ^{2} - 9} d x

\frac{\partial}{\partial x} (\frac{7}{x ^{3}})

\frac{d}{d x} (\frac{π x}{2})

\int (sin^{3} (x) cos (x)) d x

What is the integral of (3x^3+9x^2+2x+6)/(x^4+x^2) ?
The integral of (3x^3+9x^2+2x+6)/(x^4+x^2) is 2ln|x|-6/x+1/2 ln|x^2+1|+3arctan(x)+C