What is the integral of (x^2)/(x^4-6x^2-27) ?

The integral of (x^2)/(x^4-6x^2-27) is 1/(4sqrt(3))arctan(x/(sqrt(3)))-1/8 ln|x+3|+1/8 ln|x-3|+C

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integral of (x^2)/(x^4-6x^2-27)

\int \frac{x ^{2}}{x ^{4} - 6 x ^{2} - 27} d x

\frac{1}{4 3} arctan (\frac{x}{3}) - \frac{1}{8} ln ∣ x + 3 ∣ + \frac{1}{8} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x ^{2}}{x ^{4} - 6 x ^{2} - 27} d x

Take the partial fraction of

\frac{x ^{2}}{x ^{4} - 6 x ^{2} - 27} : \frac{1}{4 ( x ^{2} + 3 )} - \frac{1}{8 ( x + 3 )} + \frac{1}{8 ( x - 3 )}

= \int \frac{1}{4 ( x ^{2} + 3 )} - \frac{1}{8 ( x + 3 )} + \frac{1}{8 ( x - 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}{4 ( x ^{2} + 3 )} d x - \int \frac{1}{8 ( x + 3 )} d x + \int \frac{1}{8 ( x - 3 )} d x

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

\int \frac{1}{8 ( x + 3 )} d x = \frac{1}{8} ln ∣ x + 3 ∣

\int \frac{1}{8 ( x - 3 )} d x = \frac{1}{8} ln ∣ x - 3 ∣

= \frac{1}{4 3} arctan (\frac{x}{3}) - \frac{1}{8} ln ∣ x + 3 ∣ + \frac{1}{8} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{1}{4 3} arctan (\frac{x}{3}) - \frac{1}{8} ln ∣ x + 3 ∣ + \frac{1}{8} ln ∣ x - 3 ∣ + C

y y^{^{'}} + 36 x = 0

\frac{d}{d x} ((8 x^{2} - 16 x + 16) e^{x})

\int \frac{20 x ^{3} - 40 x ^{2} + 9}{x ^{2} - 2 x} d x

\frac{\partial}{\partial x} ((2 x + 1)^{2})

\frac{d}{d x} (4 x^{2})

What is the integral of (x^2)/(x^4-6x^2-27) ?
The integral of (x^2)/(x^4-6x^2-27) is 1/(4sqrt(3))arctan(x/(sqrt(3)))-1/8 ln|x+3|+1/8 ln|x-3|+C