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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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