What is the integral of (x^4)/(4-9x^{10)} ?

The integral of (x^4)/(4-9x^{10)} is 1/60 (ln| 3/2 x^5+1|-ln| 3/2 x^5-1|)+C

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integral of (x^4)/(4-9x^{10)}

\int \frac{x ^{4}}{4 - 9 x ^{10}} d x

\frac{1}{60} (ln \frac{3}{2} x^{5} + 1 - ln \frac{3}{2} x^{5} - 1) + C

Solution steps

\int \frac{x ^{4}}{4 - 9 x ^{10}} d x

Apply u-substitution

: \int \frac{1}{5 ( 4 - 9 u ^{2} )} d u

= \int \frac{1}{5 ( 4 - 9 u ^{2} )} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{4 - 9 u ^{2}} d u

Apply Integral Substitution

= \frac{1}{5} \cdot \int \frac{1}{6 ( - 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}{5} \cdot \frac{1}{6} \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}{5} \cdot \frac{1}{6} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= \frac{1}{5} \cdot \frac{1}{6} (\frac{ln \frac{3}{2} x ^{5} + 1}{2} - \frac{ln \frac{3}{2} x ^{5} - 1}{2})

Simplify

\frac{1}{5} \cdot \frac{1}{6} (\frac{ln \frac{3}{2} x ^{5} + 1}{2} - \frac{ln \frac{3}{2} x ^{5} - 1}{2}) : \frac{1}{60} (ln \frac{3}{2} x^{5} + 1 - ln \frac{3}{2} x^{5} - 1)

= \frac{1}{60} (ln \frac{3}{2} x^{5} + 1 - ln \frac{3}{2} x^{5} - 1)

Add a constant to the solution

= \frac{1}{60} (ln \frac{3}{2} x^{5} + 1 - ln \frac{3}{2} x^{5} - 1) + C

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What is the integral of (x^4)/(4-9x^{10)} ?
The integral of (x^4)/(4-9x^{10)} is 1/60 (ln| 3/2 x^5+1|-ln| 3/2 x^5-1|)+C