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

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

Popular >

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

x \frac{d y}{d x} = y^{2} - y

\int_{1}^{4} t^{- \frac{3}{2}} d t

x^{^{''}} + 10 x^{^{'}} + 25 = 14 e^{- 5 t}

\frac{d y}{d t} = - 4 (y - 3)

x^{4} - 50 x^{2}

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