What is the integral of 1/(3x^2+12) ?

The integral of 1/(3x^2+12) is 1/6 arctan(x/2)+C

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integral of 1/(3x^2+12)

\int \frac{1}{3 x ^{2} + 12} d x

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

Solution steps

\int \frac{1}{3 x ^{2} + 12} d x

Apply Integral Substitution

= \int \frac{1}{6 ( 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}{6} \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}{6} arctan (u)

Substitute back

u = \frac{3}{2 3} x

= \frac{1}{6} arctan (\frac{3}{2 3} x)

Simplify

\frac{1}{6} arctan (\frac{3}{2 3} x) : \frac{1}{6} arctan (\frac{x}{2})

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

Add a constant to the solution

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

\int u^{- 5} (7 - u^{5} + u^{10}) d u

\frac{d y}{d x} - 12 y = e^{9 x} y^{2}

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

x \to 0 lim (e^{x^{2}} - 1)

\frac{d y}{d x} = x^{2} - y - 2, y (0) = 1

What is the integral of 1/(3x^2+12) ?
The integral of 1/(3x^2+12) is 1/6 arctan(x/2)+C