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

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

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integral of x/(x^4+6)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 6 )} d u

= \int \frac{1}{2 ( u ^{2} + 6 )} 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}{u ^{2} + 6} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{6 ( 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}{6} \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}{6} arctan (v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (y (1 - e^{- x}))

\int 2 x + 4 sin (x) d x

\int_{0}^{\infty} x \cdot e^{- 2 x} d x

\frac{d y}{d x} = y \cdot tan (x) + 1

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

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