What is the integral of (x^3)/(1-x^8) ?

The integral of (x^3)/(1-x^8) is 1/8 (ln|x^4+1|-ln|x^4-1|)+C

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integral of (x^3)/(1-x^8)

\int \frac{x ^{3}}{1 - x ^{8}} d x

\frac{1}{8} (ln x^{4} + 1 - ln x^{4} - 1) + C

Solution steps

\int \frac{x ^{3}}{1 - x ^{8}} d x

Factor

1 - x^{8} : - (- 1 + x^{8})

= \int \frac{x ^{3}}{- ( - 1 + x ^{8} )} d x

Take the constant out:

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

= - \int \frac{x ^{3}}{- 1 + x ^{8}} d x

Apply u-substitution

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

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

Factor

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

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

Take the constant out:

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

= - \frac{1}{4} (- \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}

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

Substitute back

u = x^{4}

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

Simplify

- \frac{1}{4} (- (\frac{ln x ^{4} + 1}{2} - \frac{ln x ^{4} - 1}{2})) : \frac{1}{8} (ln x^{4} + 1 - ln x^{4} - 1)

= \frac{1}{8} (ln x^{4} + 1 - ln x^{4} - 1)

Add a constant to the solution

= \frac{1}{8} (ln x^{4} + 1 - ln x^{4} - 1) + C

\frac{1 - 2 ln ( x )}{x ^{3}}

\int \frac{64 - x ^{2}}{x ^{2}} d x

\int_{1}^{3} \frac{9}{x ^{2}} d x

\int e^{- 4 x} sin (5 x) d x

f (x) = \frac{110}{x ^{12}}

What is the integral of (x^3)/(1-x^8) ?
The integral of (x^3)/(1-x^8) is 1/8 (ln|x^4+1|-ln|x^4-1|)+C