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

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

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integral of x/(2+x^4)

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{2 ( 1 + v ^{2} )} 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}{1 + v ^{2}} d v

Use the common integral:

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

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

Add a constant to the solution

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

\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{7}{y ^{4}}) (y + 5 y^{3}))

\frac{1}{( s ^{2} - 2 s )}

\int_{0}^{5} \frac{1}{x ^{2} - 6 x + 5} d x

x \to 3 + lim (ln x^{2} - 9)

\frac{d ^{3}}{d x ^{3}} (6 x + 8)

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