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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{4 u 1 - u} d u

= \int \frac{1}{4 u 1 - u} d u

Take the constant out:

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

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

Apply u-substitution

: \int - \frac{2}{- v ^{2} + 1} d v

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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