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

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

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integral of 4/(y^4-1)

\int \frac{4}{y ^{4} - 1} d y

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

Solution steps

\int \frac{4}{y ^{4} - 1} d y

Take the constant out:

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

= 4 \cdot \int \frac{1}{y ^{4} - 1} d y

Take the partial fraction of

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

= 4 \cdot \int - \frac{1}{2 ( y ^{2} + 1 )} - \frac{1}{4 ( y + 1 )} + \frac{1}{4 ( y - 1 )} d y

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 ( y ^{2} + 1 )} d y - \int \frac{1}{4 ( y + 1 )} d y + \int \frac{1}{4 ( y - 1 )} d y)

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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