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

The integral of x/(16+x^4) is 1/8 arctan((x^2)/4)+C

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integral of x/(16+x^4)

\int \frac{x}{16 + x ^{4}} d x

\frac{1}{8} arctan (\frac{x ^{2}}{4}) + C

Solution steps

\int \frac{x}{16 + x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 ( 16 + u ^{2} )} d u

= \int \frac{1}{2 ( 16 + 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}{16 + u ^{2}} d u

Apply Integral Substitution

= \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

= \frac{1}{2} \cdot \frac{1}{4} \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}{4} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{4} arctan (\frac{x ^{2}}{4})

Simplify

\frac{1}{2} \cdot \frac{1}{4} arctan (\frac{x ^{2}}{4}) : \frac{1}{8} arctan (\frac{x ^{2}}{4})

= \frac{1}{8} arctan (\frac{x ^{2}}{4})

Add a constant to the solution

= \frac{1}{8} arctan (\frac{x ^{2}}{4}) + C

x^{2} e^{x^{3}}

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What is the integral of x/(16+x^4) ?
The integral of x/(16+x^4) is 1/8 arctan((x^2)/4)+C