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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

{t^{2} - 3 t - 2 e^{- t} sin (3 t)}

7 x^{2} cos (x) cot (x)

x \to 0 lim (e^{x} + 3 x \frac{1}{x})

\frac{d}{d t} (4 e^{\frac{t}{2}})

arcsin (6 x + 1)

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