What is the integral of (x^2+x+3)/(x^4+6x^2+9) ?

The integral of (x^2+x+3)/(x^4+6x^2+9) is 1/(sqrt(3))arctan(x/(sqrt(3)))-1/(2(x^2+3))+C

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integral of (x^2+x+3)/(x^4+6x^2+9)

\int \frac{x ^{2} + x + 3}{x ^{4} + 6 x ^{2} + 9} d x

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

Solution steps

\int \frac{x ^{2} + x + 3}{x ^{4} + 6 x ^{2} + 9} d x

Take the partial fraction of

\frac{x ^{2} + x + 3}{x ^{4} + 6 x ^{2} + 9} : \frac{1}{x ^{2} + 3} + \frac{x}{( x ^{2} + 3 ) ^{2}}

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

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

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

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

Add a constant to the solution

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

\int \frac{1}{- x ^{2}} d x

f (x) = \frac{e ^{x}}{x}, (1, e)

\frac{d}{d x} (\frac{x ^{2} - 6 x + 5}{x ^{2} + 7})

\int_{0}^{3} (3 x^{2} + x - 2) d x

What is the integral of (x^2+x+3)/(x^4+6x^2+9) ?
The integral of (x^2+x+3)/(x^4+6x^2+9) is 1/(sqrt(3))arctan(x/(sqrt(3)))-1/(2(x^2+3))+C