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

The integral of (x^4)/(x^4+5x^2+4) is x+1/3 arctan(x)-8/3 arctan(x/2)+C

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

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

x + \frac{1}{3} arctan (x) - \frac{8}{3} arctan (\frac{x}{2}) + C

Solution steps

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

Take the partial fraction of

\frac{x ^{4}}{x ^{4} + 5 x ^{2} + 4} : 1 + \frac{1}{3 ( x ^{2} + 1 )} - \frac{16}{3 ( x ^{2} + 4 )}

= \int 1 + \frac{1}{3 ( x ^{2} + 1 )} - \frac{16}{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 1 d x + \int \frac{1}{3 ( x ^{2} + 1 )} d x - \int \frac{16}{3 ( x ^{2} + 4 )} d x

\int 1 d x = x

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

\int \frac{16}{3 ( x ^{2} + 4 )} d x = \frac{8}{3} arctan (\frac{x}{2})

= x + \frac{1}{3} arctan (x) - \frac{8}{3} arctan (\frac{x}{2})

Add a constant to the solution

= x + \frac{1}{3} arctan (x) - \frac{8}{3} arctan (\frac{x}{2}) + C

What is the integral of (x^4)/(x^4+5x^2+4) ?
The integral of (x^4)/(x^4+5x^2+4) is x+1/3 arctan(x)-8/3 arctan(x/2)+C