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

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

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integral of x/(4+x^4)

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (\frac{7}{x ^{2} + y ^{2}})

\frac{( x ^{2} + 32 x )}{8}, at x = 4

(t + 2)^{\frac{2}{3}} (2 t^{2} - 3)^{3}

\frac{d}{d x} (f (x) \cdot \frac{1}{g ( x )})

\frac{y ^{^{'}} - e ^{- t} + 6}{y} = - 6

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