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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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