What is the integral of x/(x^4+a^4) ?

The integral of x/(x^4+a^4) is 1/(2a^2)arctan((x^2)/(a^2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x/(x^4+a^4)

\int \frac{x}{x ^{4} + a ^{4}} d x

\frac{1}{2 a ^{2}} arctan (\frac{x ^{2}}{a ^{2}}) + C

Solution steps

\int \frac{x}{x ^{4} + a ^{4}} d x

Apply u-substitution

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{2} + a ^{4}} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{a ^{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}{2} \cdot \frac{1}{a ^{2}} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{2} \cdot \frac{1}{a ^{2}} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{a ^{2}} arctan (\frac{x ^{2}}{a ^{2}})

Simplify

\frac{1}{2} \cdot \frac{1}{a ^{2}} arctan (\frac{x ^{2}}{a ^{2}}) : \frac{1}{2 a ^{2}} arctan (\frac{x ^{2}}{a ^{2}})

= \frac{1}{2 a ^{2}} arctan (\frac{x ^{2}}{a ^{2}})

Add a constant to the solution

= \frac{1}{2 a ^{2}} arctan (\frac{x ^{2}}{a ^{2}}) + C

What is the integral of x/(x^4+a^4) ?
The integral of x/(x^4+a^4) is 1/(2a^2)arctan((x^2)/(a^2))+C