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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{1 - u ^{2}} 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

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