What is the integral of (x^2)/((4+x^6)) ?

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

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integral of (x^2)/((4+x^6))

integral

\frac{x ^{2}}{( 4 + x ^{6} )}

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

Solution steps

\int \frac{x ^{2}}{4 + x ^{6}} d x

Apply u-substitution

: \int \frac{1}{3 ( 4 + u ^{2} )} d u

= \int \frac{1}{3 ( 4 + u ^{2} )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{4 + u ^{2}} d u

Apply Integral Substitution

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{1}{x} + \frac{1}{x ^{3}}

\frac{( 3 x - 13 )}{( x ^{2} - 12 + 45 ) ^{2}}

4 x^{3} + 5 x^{4} + 3

\frac{( x + 5 )}{( - 3 x )}

sin^{3} (x) - 1

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