What is the integral of (x^3)/(x^8-4) ?

The integral of (x^3)/(x^8-4) is -1/16 (ln|(x^4)/2+1|-ln|(x^4)/2-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^3)/(x^8-4)

\int \frac{x ^{3}}{x ^{8} - 4} d x

- \frac{1}{16} (ln \frac{x ^{4}}{2} + 1 - ln \frac{x ^{4}}{2} - 1) + C

Solution steps

\int \frac{x ^{3}}{x ^{8} - 4} d x

Factor

x^{8} - 4 : - (- x^{8} + 4)

= \int \frac{x ^{3}}{- ( - x ^{8} + 4 )} d x

Take the constant out:

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

= - \int \frac{x ^{3}}{- x ^{8} + 4} d x

Apply u-substitution

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

Apply Integral Substitution

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

Substitute back

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

Simplify

- \frac{1}{4} \cdot \frac{1}{2} \frac{ln \frac{x ^{4}}{2} + 1}{2} - \frac{ln \frac{x ^{4}}{2} - 1}{2} : - \frac{1}{16} (ln \frac{x ^{4}}{2} + 1 - ln \frac{x ^{4}}{2} - 1)

= - \frac{1}{16} (ln \frac{x ^{4}}{2} + 1 - ln \frac{x ^{4}}{2} - 1)

Add a constant to the solution

= - \frac{1}{16} (ln \frac{x ^{4}}{2} + 1 - ln \frac{x ^{4}}{2} - 1) + C

What is the integral of (x^3)/(x^8-4) ?
The integral of (x^3)/(x^8-4) is -1/16 (ln|(x^4)/2+1|-ln|(x^4)/2-1|)+C