What is the integral of (4x^2+4x+4)/((x^2+1)^2) ?

The integral of (4x^2+4x+4)/((x^2+1)^2) is 4arctan(x)-2/(x^2+1)+C

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integral of (4x^2+4x+4)/((x^2+1)^2)

\int \frac{4 x ^{2} + 4 x + 4}{( x ^{2} + 1 ) ^{2}} d x

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

Solution steps

\int \frac{4 x ^{2} + 4 x + 4}{( x ^{2} + 1 ) ^{2}} d x

Expand

\frac{4 x ^{2} + 4 x + 4}{( x ^{2} + 1 ) ^{2}} : \frac{4 x ^{2}}{( x ^{2} + 1 ) ^{2}} + \frac{4 x}{( x ^{2} + 1 ) ^{2}} + \frac{4}{( x ^{2} + 1 ) ^{2}}

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{4 x ^{2}}{( x ^{2} + 1 ) ^{2}} d x + \int \frac{4 x}{( x ^{2} + 1 ) ^{2}} d x + \int \frac{4}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{4 x ^{2}}{( x ^{2} + 1 ) ^{2}} d x = 4 (\frac{1}{2} arctan (x) - \frac{x}{2 ( 1 + x ^{2} )})

\int \frac{4 x}{( x ^{2} + 1 ) ^{2}} d x = - \frac{2}{x ^{2} + 1}

\int \frac{4}{( x ^{2} + 1 ) ^{2}} d x = 2 (arctan (x) + \frac{x}{1 + x ^{2}})

= 4 (\frac{1}{2} arctan (x) - \frac{x}{2 ( 1 + x ^{2} )}) - \frac{2}{x ^{2} + 1} + 2 (arctan (x) + \frac{x}{1 + x ^{2}})

Simplify

4 (\frac{1}{2} arctan (x) - \frac{x}{2 ( 1 + x ^{2} )}) - \frac{2}{x ^{2} + 1} + 2 (arctan (x) + \frac{x}{1 + x ^{2}}) : 4 arctan (x) - \frac{2}{x ^{2} + 1}

= 4 arctan (x) - \frac{2}{x ^{2} + 1}

Add a constant to the solution

= 4 arctan (x) - \frac{2}{x ^{2} + 1} + C

What is the integral of (4x^2+4x+4)/((x^2+1)^2) ?
The integral of (4x^2+4x+4)/((x^2+1)^2) is 4arctan(x)-2/(x^2+1)+C