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

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

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

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

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

Solution steps

\int \frac{2 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

= 2 \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 )}

= 2 \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

= 2 (\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 ∣

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} + \frac{1}{2} y^{^{'}} + \frac{1}{16} y = 0, y (0) = 5, y^{^{'}} (0) = 0

x \to 0 lim (\frac{x ^{3}}{x ^{6}})

\frac{\partial}{\partial y} (x^{2} + 2 y^{2} - x)

\int \frac{2 x - 3}{4 x - x ^{2}} d x

8

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