What is the integral of (3w)/(16w^4-81) ?

The integral of (3w)/(16w^4-81) is -1/48 (ln| 4/9 w^2+1|-ln| 4/9 w^2-1|)+C

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integral of (3w)/(16w^4-81)

\int \frac{3 w}{16 w ^{4} - 81} d w

- \frac{1}{48} (ln \frac{4}{9} w^{2} + 1 - ln \frac{4}{9} w^{2} - 1) + C

Solution steps

\int \frac{3 w}{16 w ^{4} - 81} d w

Take the constant out:

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

= 3 \cdot \int \frac{w}{16 w ^{4} - 81} d w

Factor

16 w^{4} - 81 : - (- 16 w^{4} + 81)

= 3 \cdot \int \frac{w}{- ( - 16 w ^{4} + 81 )} d w

Take the constant out:

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

= 3 (- \int \frac{w}{- 16 w ^{4} + 81} d w)

Apply u-substitution

: \int \frac{1}{2 ( - 16 u ^{2} + 81 )} d u

= 3 (- \int \frac{1}{2 ( - 16 u ^{2} + 81 )} d u)

Take the constant out:

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

= 3 (- \frac{1}{2} \cdot \int \frac{1}{- 16 u ^{2} + 81} d u)

Apply Integral Substitution

= 3 (- \frac{1}{2} \cdot \int \frac{1}{36 ( - v ^{2} + 1 )} d v)

Take the constant out:

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

= 3 (- \frac{1}{2} \cdot \frac{1}{36} \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}

= 3 (- \frac{1}{2} \cdot \frac{1}{36} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 3 (- \frac{1}{2} \cdot \frac{1}{36} (\frac{ln \frac{4}{9} w ^{2} + 1}{2} - \frac{ln \frac{4}{9} w ^{2} - 1}{2}))

Simplify

3 (- \frac{1}{2} \cdot \frac{1}{36} (\frac{ln \frac{4}{9} w ^{2} + 1}{2} - \frac{ln \frac{4}{9} w ^{2} - 1}{2})) : - \frac{1}{48} (ln \frac{4}{9} w^{2} + 1 - ln \frac{4}{9} w^{2} - 1)

= - \frac{1}{48} (ln \frac{4}{9} w^{2} + 1 - ln \frac{4}{9} w^{2} - 1)

Add a constant to the solution

= - \frac{1}{48} (ln \frac{4}{9} w^{2} + 1 - ln \frac{4}{9} w^{2} - 1) + C

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What is the integral of (3w)/(16w^4-81) ?
The integral of (3w)/(16w^4-81) is -1/48 (ln| 4/9 w^2+1|-ln| 4/9 w^2-1|)+C