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

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

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

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

\frac{5}{6} ln x^{2} + 1 + arctan (x) - \frac{5}{6} ln x^{2} + 4 + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 5 x + 4}{x ^{4} + 5 x ^{2} + 4} : \frac{5 x + 3}{3 ( x ^{2} + 1 )} - \frac{5 x}{3 ( x ^{2} + 4 )}

= \int \frac{5 x + 3}{3 ( x ^{2} + 1 )} - \frac{5 x}{3 ( x ^{2} + 4 )} 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{5 x + 3}{3 ( x ^{2} + 1 )} d x - \int \frac{5 x}{3 ( x ^{2} + 4 )} d x

\int \frac{5 x + 3}{3 ( x ^{2} + 1 )} d x = \frac{5}{6} ln x^{2} + 1 + arctan (x)

\int \frac{5 x}{3 ( x ^{2} + 4 )} d x = \frac{5}{6} ln x^{2} + 4

= \frac{5}{6} ln x^{2} + 1 + arctan (x) - \frac{5}{6} ln x^{2} + 4

Add a constant to the solution

= \frac{5}{6} ln x^{2} + 1 + arctan (x) - \frac{5}{6} ln x^{2} + 4 + C

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