What is the integral of (-6x^2)/(x^4-1) ?

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

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integral of (-6x^2)/(x^4-1)

\int \frac{- 6 x ^{2}}{x ^{4} - 1} d x

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

Solution steps

\int \frac{- 6 x ^{2}}{x ^{4} - 1} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 6 \cdot \int \frac{x ^{2}}{x ^{4} - 1} d x

Take the partial fraction of

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

= - 6 \cdot \int \frac{1}{2 ( x ^{2} + 1 )} - \frac{1}{4 ( x + 1 )} + \frac{1}{4 ( x - 1 )} 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

= - 6 (\int \frac{1}{2 ( x ^{2} + 1 )} d x - \int \frac{1}{4 ( x + 1 )} d x + \int \frac{1}{4 ( x - 1 )} d x)

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

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

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

Add a constant to the solution

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

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

\int x ln^{2} (x) d x

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{23}{2}}} d x

x \to 10 - lim (ln (100 - x^{2}))

\frac{\partial}{\partial y} (2 y^{2})

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