What is the integral of (x^3+x^2+2x+3)/(x^4+5x^2+6) ?

The integral of (x^3+x^2+2x+3)/(x^4+5x^2+6) is 1/(sqrt(2))arctan(x/(sqrt(2)))+1/2 ln|x^2+3|+C

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integral of (x^3+x^2+2x+3)/(x^4+5x^2+6)

\int \frac{x ^{3} + x ^{2} + 2 x + 3}{x ^{4} + 5 x ^{2} + 6} d x

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

Solution steps

\int \frac{x ^{3} + x ^{2} + 2 x + 3}{x ^{4} + 5 x ^{2} + 6} d x

Take the partial fraction of

\frac{x ^{3} + x ^{2} + 2 x + 3}{x ^{4} + 5 x ^{2} + 6} : \frac{1}{x ^{2} + 2} + \frac{x}{x ^{2} + 3}

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

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

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

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

Add a constant to the solution

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

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What is the integral of (x^3+x^2+2x+3)/(x^4+5x^2+6) ?
The integral of (x^3+x^2+2x+3)/(x^4+5x^2+6) is 1/(sqrt(2))arctan(x/(sqrt(2)))+1/2 ln|x^2+3|+C