What is the integral of (5x)/(x^4-16) ?

The integral of (5x)/(x^4-16) is -5/16 (ln|(x^2)/4+1|-ln|(x^2)/4-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (5x)/(x^4-16)

\int \frac{5 x}{x ^{4} - 16} d x

- \frac{5}{16} (ln \frac{x ^{2}}{4} + 1 - ln \frac{x ^{2}}{4} - 1) + C

Solution steps

\int \frac{5 x}{x ^{4} - 16} d x

Take the constant out:

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

= 5 \cdot \int \frac{x}{x ^{4} - 16} d x

Apply u-substitution

= 5 \cdot \int \frac{1}{2 ( u ^{2} - 16 )} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} - 16} d u

Apply Integral Substitution

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{4 ( v ^{2} - 1 )} d v

Take the constant out:

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

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

Factor

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

= 5 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \frac{1}{4} (- \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}

= 5 \cdot \frac{1}{2} \cdot \frac{1}{4} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 5 \cdot \frac{1}{2} \cdot \frac{1}{4} - \frac{ln \frac{x ^{2}}{4} + 1}{2} - \frac{ln \frac{x ^{2}}{4} - 1}{2}

Simplify

5 \cdot \frac{1}{2} \cdot \frac{1}{4} - \frac{ln \frac{x ^{2}}{4} + 1}{2} - \frac{ln \frac{x ^{2}}{4} - 1}{2} : - \frac{5}{16} (ln \frac{x ^{2}}{4} + 1 - ln \frac{x ^{2}}{4} - 1)

= - \frac{5}{16} (ln \frac{x ^{2}}{4} + 1 - ln \frac{x ^{2}}{4} - 1)

Add a constant to the solution

= - \frac{5}{16} (ln \frac{x ^{2}}{4} + 1 - ln \frac{x ^{2}}{4} - 1) + C

What is the integral of (5x)/(x^4-16) ?
The integral of (5x)/(x^4-16) is -5/16 (ln|(x^2)/4+1|-ln|(x^2)/4-1|)+C