What is the integral of 1/(2x^2+8) ?

The integral of 1/(2x^2+8) is 1/4 arctan(x/2)+C

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integral of 1/(2x^2+8)

\int \frac{1}{2 x ^{2} + 8} d x

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

Solution steps

\int \frac{1}{2 x ^{2} + 8} d x

Apply Integral Substitution

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

Use the common integral:

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

= \frac{1}{4} arctan (u)

Substitute back

u = \frac{2}{2 2} x

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

Simplify

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

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

Add a constant to the solution

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

x \to 0 - lim (\frac{1}{x} + \frac{1}{∣ x ∣})

\frac{d y}{d x} = \frac{x}{6 y}

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

\frac{d}{d x} (e^{4 e^{x}})

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

What is the integral of 1/(2x^2+8) ?
The integral of 1/(2x^2+8) is 1/4 arctan(x/2)+C