What is the integral of (5x^2+7x+5)/((x^2+1)^2) ?

The integral of (5x^2+7x+5)/((x^2+1)^2) is 5arctan(x)-7/(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 (5x^2+7x+5)/((x^2+1)^2)

\int \frac{5 x ^{2} + 7 x + 5}{( x ^{2} + 1 ) ^{2}} d x

5 arctan (x) - \frac{7}{2 ( x ^{2} + 1 )} + C

Solution steps

\int \frac{5 x ^{2} + 7 x + 5}{( x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{5 x ^{2} + 7 x + 5}{( x ^{2} + 1 ) ^{2}} : \frac{5}{x ^{2} + 1} + \frac{7 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{5}{x ^{2} + 1} + \frac{7 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{5}{x ^{2} + 1} d x + \int \frac{7 x}{( x ^{2} + 1 ) ^{2}} d x

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

\int \frac{7 x}{( x ^{2} + 1 ) ^{2}} d x = - \frac{7}{2 ( x ^{2} + 1 )}

= 5 arctan (x) - \frac{7}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 5 arctan (x) - \frac{7}{2 ( x ^{2} + 1 )} + C

\frac{d}{d x} ((x + 5) (x - 1) (x - 7))

\int sin^{5} (x co) s^{6} x d x

f (x) = \frac{x ^{3} - 7 x ^{2} + 7}{x ^{2}}

\frac{\partial}{\partial x} (y^{6} sin (2 x))

x \to \infty lim (- 3 x)

What is the integral of (5x^2+7x+5)/((x^2+1)^2) ?
The integral of (5x^2+7x+5)/((x^2+1)^2) is 5arctan(x)-7/(2(x^2+1))+C