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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{16}{x ^{2}} - \frac{4}{x})

f (x) = (a^{\frac{1}{x}} + b^{\frac{1}{x}})^{x}

\int \frac{x ^{3}}{5 x ^{2} + 2} d x

\frac{8}{s ^{2} + 9}

\frac{d ^{2}}{d x ^{2}} (\frac{x - 4}{3 x - x ^{2}})

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