What is the integral of x/(x^4+x^2+1) ?

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

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integral of x/(x^4+x^2+1)

\int \frac{x}{x ^{4} + x ^{2} + 1} d x

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

Solution steps

\int \frac{x}{x ^{4} + x ^{2} + 1} d x

Apply u-substitution

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

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{u ( u + u + 1 )} = u^{- \frac{1}{2}} \frac{1}{( + u + 1 )}

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

Simplify

u^{- \frac{1}{2}} \frac{1}{u + 1} : \frac{1}{u ( u + 1 )}

= \frac{1}{4} \cdot \int \frac{1}{u ( u + 1 )} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 1} d v

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{( x ^{2} - 1 ) ^{4}}{x ^{2} + 1})

x \to \infty lim (x^{2} - 9)

(x^{2} - 60) (2 x - 3)

\int \frac{x - 4}{x + 2} d x

\int (2 t + 1) d t

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