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

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

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integral of x/(x^4-1)

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

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

Take the constant out:

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

= \frac{1}{2} (- \int \frac{1}{- u ^{2} + 1} d u)

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

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

Substitute back

u = x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (- 2 e^{- x} cos (x))

y^{^{'}} + 2 y = 0, y (0) = 1

4 y^{^{''}} + 4 y^{^{'}} + 17 = 0

\frac{d}{d x} (\frac{x + 7}{4 x ^{2} e ^{x}})

\int arcsin (\frac{4}{3} x + 5) d x

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