What is the integral of x/(81x^4-1) ?

The integral of x/(81x^4-1) is -1/36 (ln|9x^2+1|-ln|9x^2-1|)+C

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integral of x/(81x^4-1)

integral

\frac{x}{81 x ^{4} - 1}

- \frac{1}{36} (ln 9 x^{2} + 1 - ln 9 x^{2} - 1) + C

Solution steps

\int \frac{x}{81 x ^{4} - 1} d x

Apply u-substitution

= \int \frac{1}{2 ( 81 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}{81 u ^{2} - 1} d u

Factor

81 u^{2} - 1 : - (- 81 u^{2} + 1)

= \frac{1}{2} \cdot \int \frac{1}{- ( - 81 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}{- 81 u ^{2} + 1} d u)

Apply Integral Substitution

= \frac{1}{2} (- \int \frac{1}{9 ( - 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}{2} (- \frac{1}{9} \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}{2} (- \frac{1}{9} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= \frac{1}{2} (- \frac{1}{9} (\frac{ln 9 x ^{2} + 1}{2} - \frac{ln 9 x ^{2} - 1}{2}))

Simplify

\frac{1}{2} (- \frac{1}{9} (\frac{ln 9 x ^{2} + 1}{2} - \frac{ln 9 x ^{2} - 1}{2})) : - \frac{1}{36} (ln 9 x^{2} + 1 - ln 9 x^{2} - 1)

= - \frac{1}{36} (ln 9 x^{2} + 1 - ln 9 x^{2} - 1)

Add a constant to the solution

= - \frac{1}{36} (ln 9 x^{2} + 1 - ln 9 x^{2} - 1) + C

in t e g r a l \frac{1}{( 4 - x ^{2} )}

in t e g r a l \frac{1}{( x ^{2} + x + 1 )}

in t e g r a l cos^{3} (x) - cos (x)

in t e g r a l sin (2 - 4 x)

in t e g r a l \frac{x}{( x ^{2} + a ^{2} )}

What is the integral of x/(81x^4-1) ?
The integral of x/(81x^4-1) is -1/36 (ln|9x^2+1|-ln|9x^2-1|)+C