What is the integral of (x^2+3x+2)/(x(x^2+1)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2+3x+2)/(x(x^2+1))

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Add similar elements:

\frac{1}{2} ln x^{2} + 1 - ln x^{2} + 1 = - \frac{1}{2} ln x^{2} + 1

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

Add a constant to the solution

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

\int \frac{x ^{2} + 7 x - 6}{( x + 1 ) ( x ^{2} - 4 x + 4 )} d x

\int \frac{x}{( x - 6 )} d x

\int \frac{7 x - 9}{x ^{2} - 4 x - 21} d x

\int \frac{y ^{2}}{9 - 8 y - y ^{2}} d y

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