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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

Simplify

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

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

Add a constant to the solution

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

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