What is the integral of x/(9+x^4) ?

The integral of x/(9+x^4) is 1/6 arctan((x^2)/3)+C

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integral of x/(9+x^4)

\int \frac{x}{9 + x ^{4}} d x

\frac{1}{6} arctan (\frac{x ^{2}}{3}) + C

Solution steps

\int \frac{x}{9 + x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 ( 9 + u ^{2} )} d u

= \int \frac{1}{2 ( 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}{2} \cdot \int \frac{1}{9 + u ^{2}} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{3 ( 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}{3} \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}{3} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{3} arctan (\frac{x ^{2}}{3})

Simplify

\frac{1}{2} \cdot \frac{1}{3} arctan (\frac{x ^{2}}{3}) : \frac{1}{6} arctan (\frac{x ^{2}}{3})

= \frac{1}{6} arctan (\frac{x ^{2}}{3})

Add a constant to the solution

= \frac{1}{6} arctan (\frac{x ^{2}}{3}) + C

ln (16 sin^{2} (x))

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What is the integral of x/(9+x^4) ?
The integral of x/(9+x^4) is 1/6 arctan((x^2)/3)+C