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

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

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

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

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

Solution steps

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

Apply Integral Substitution

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

Substitute back

u = \frac{2}{2} x

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{a ^{2} sin ^{2} ( x ) + b ^{2} cos ^{2} ( x )} d x

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

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + a ^{2}})

\frac{x ^{2}}{3 ( 1 - x )}

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