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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= 4 \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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= 4 \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

= 4 \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}

= 4 \cdot \frac{1}{2} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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