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

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

integral of 1/(x^2+12)

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

\frac{1}{2 3} arctan (\frac{x}{2 3}) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{2 3 ( 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 3} \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 3} arctan (u)

Substitute back

u = \frac{x}{2 3}

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

Add a constant to the solution

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

3 x, 9 x^{2}

n = 1 \sum \infty \frac{2 n}{1 1 ^{n}}

\frac{d}{d x} (\frac{4 2}{3} x^{\frac{3}{2}} - 1)

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

\int 3 x^{2} y + 4 x y^{2} d x

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