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

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

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

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

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

Solution steps

\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 )}

= \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

= \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 ∣

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

Add a constant to the solution

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

\int \frac{3}{x ^{3}} d x

\int_{- 3}^{5} (y^{3} - 4 y) d y

\int_{0}^{2} 3 d x

\int_{- \infty}^{0} e^{4 x} d x

x \to 0 lim (e^{3 x})

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