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

The integral of (x^2+x+1)/((2x+1)(x^2+1)) is 3/10 ln|2x+1|+1/5 (1/2 ln|x^2+1|+2arctan(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{3}{10} ln ∣ 2 x + 1 ∣ + \frac{1}{5} (\frac{1}{2} ln x^{2} + 1 + 2 arctan (x)) + C

Solution steps

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

Take the partial fraction of

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

= \int \frac{3}{5 ( 2 x + 1 )} + \frac{x + 2}{5 ( 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{3}{5 ( 2 x + 1 )} d x + \int \frac{x + 2}{5 ( x ^{2} + 1 )} d x

\int \frac{3}{5 ( 2 x + 1 )} d x = \frac{3}{10} ln ∣ 2 x + 1 ∣

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

= \frac{3}{10} ln ∣ 2 x + 1 ∣ + \frac{1}{5} (\frac{1}{2} ln x^{2} + 1 + 2 arctan (x))

Add a constant to the solution

= \frac{3}{10} ln ∣ 2 x + 1 ∣ + \frac{1}{5} (\frac{1}{2} ln x^{2} + 1 + 2 arctan (x)) + C

What is the integral of (x^2+x+1)/((2x+1)(x^2+1)) ?
The integral of (x^2+x+1)/((2x+1)(x^2+1)) is 3/10 ln|2x+1|+1/5 (1/2 ln|x^2+1|+2arctan(x))+C