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

The integral of x/(5-4x^2-x^4) is -1/12 ln|x+1|-1/12 ln|x-1|+1/12 ln|x^2+5|+C

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

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

- \frac{1}{12} ln ∣ x + 1 ∣ - \frac{1}{12} ln ∣ x - 1 ∣ + \frac{1}{12} ln x^{2} + 5 + C

Solution steps

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

Take the partial fraction of

\frac{x}{5 - 4 x ^{2} - x ^{4}} : - \frac{1}{12 ( x + 1 )} - \frac{1}{12 ( x - 1 )} + \frac{x}{6 ( x ^{2} + 5 )}

= \int - \frac{1}{12 ( x + 1 )} - \frac{1}{12 ( x - 1 )} + \frac{x}{6 ( x ^{2} + 5 )} 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}{12 ( x + 1 )} d x - \int \frac{1}{12 ( x - 1 )} d x + \int \frac{x}{6 ( x ^{2} + 5 )} d x

\int \frac{1}{12 ( x + 1 )} d x = \frac{1}{12} ln ∣ x + 1 ∣

\int \frac{1}{12 ( x - 1 )} d x = \frac{1}{12} ln ∣ x - 1 ∣

\int \frac{x}{6 ( x ^{2} + 5 )} d x = \frac{1}{12} ln x^{2} + 5

= - \frac{1}{12} ln ∣ x + 1 ∣ - \frac{1}{12} ln ∣ x - 1 ∣ + \frac{1}{12} ln x^{2} + 5

Add a constant to the solution

= - \frac{1}{12} ln ∣ x + 1 ∣ - \frac{1}{12} ln ∣ x - 1 ∣ + \frac{1}{12} ln x^{2} + 5 + C

\frac{12 e ^{- s}}{s ( 10 s + 1 )}

\int_{0}^{8} \frac{3}{8 - x} d x

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

\frac{d}{d x} ((tan (x))^{s i n (x)})

\frac{d}{d x} (- 2 e^{2 x})

What is the integral of x/(5-4x^2-x^4) ?
The integral of x/(5-4x^2-x^4) is -1/12 ln|x+1|-1/12 ln|x-1|+1/12 ln|x^2+5|+C