What is the integral of (x^3)/(x^8+16) ?

The integral of (x^3)/(x^8+16) is 1/16 arctan((x^4)/4)+C

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integral of (x^3)/(x^8+16)

\int \frac{x ^{3}}{x ^{8} + 16} d x

\frac{1}{16} arctan (\frac{x ^{4}}{4}) + C

Solution steps

\int \frac{x ^{3}}{x ^{8} + 16} d x

Apply u-substitution

: \int \frac{1}{4 ( u ^{2} + 16 )} d u

= \int \frac{1}{4 ( u ^{2} + 16 )} 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} + 16} d u

Apply Integral Substitution

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

Substitute back

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

Simplify

\frac{1}{4} \cdot \frac{1}{4} arctan (\frac{x ^{4}}{4}) : \frac{1}{16} arctan (\frac{x ^{4}}{4})

= \frac{1}{16} arctan (\frac{x ^{4}}{4})

Add a constant to the solution

= \frac{1}{16} arctan (\frac{x ^{4}}{4}) + C

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What is the integral of (x^3)/(x^8+16) ?
The integral of (x^3)/(x^8+16) is 1/16 arctan((x^4)/4)+C