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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= - 2 \cdot \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

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

Factor

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

= - 2 \cdot \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

= - 2 \cdot \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}

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

Substitute back

u = x^{2}

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

Simplify

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

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

Add a constant to the solution

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

x \to \frac{π}{3} lim (cos (x))

x \to 1 lim (\frac{- x + 1}{x - 1})

\int 12 x^{2} (2 x^{3} + 3)^{7} d x

\int \frac{x + 5}{x} d x

\int \frac{26 x ^{3} - 52 x ^{2} + 8}{x ^{2} - 2 x} d x

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