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

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

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

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

2 arctan (x) - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{4 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

= 4 \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 )}

= 4 \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

= 4 (\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 ∣

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

Simplify

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

= 2 arctan (x) - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣

Add a constant to the solution

= 2 arctan (x) - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

\int \frac{x ^{3}}{x ^{4} + 12} d x

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

\int \frac{2}{x ^{2} + 25} d x

\int 4 x^{3} sin (x^{4}) d x

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