What is the integral of (7x^2+3x+7)/((x^2+1)^2) ?

The integral of (7x^2+3x+7)/((x^2+1)^2) is 7arctan(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 (7x^2+3x+7)/((x^2+1)^2)

\int \frac{7 x ^{2} + 3 x + 7}{( x ^{2} + 1 ) ^{2}} d x

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

Solution steps

\int \frac{7 x ^{2} + 3 x + 7}{( x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{7 x ^{2} + 3 x + 7}{( x ^{2} + 1 ) ^{2}} : \frac{7}{x ^{2} + 1} + \frac{3 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{7}{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{7}{x ^{2} + 1} d x + \int \frac{3 x}{( x ^{2} + 1 ) ^{2}} d x

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

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

= 7 arctan (x) - \frac{3}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 7 arctan (x) - \frac{3}{2 ( x ^{2} + 1 )} + C

f (t) = 5 cos (8 t) + 3 e^{- 4} sin (3 t)

x \to 0 lim (ln (x) - x)

\frac{\partial}{\partial x} (y + z)

\frac{d}{d x} (cot (x) + arctan (x + 1))

\frac{d}{d x} ((x^{3} + 1)^{4})

What is the integral of (7x^2+3x+7)/((x^2+1)^2) ?
The integral of (7x^2+3x+7)/((x^2+1)^2) is 7arctan(x)-3/(2(x^2+1))+C