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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

\frac{\partial}{\partial v} (2 v cos (u))

0.882 A^{0.842}

- 8 t^{2}

f (x) = x^{\frac{1}{2}}

x \to 0 lim (\frac{1 - cos ( x )}{x})

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