What is the integral of 1/(36+x^2) ?

The integral of 1/(36+x^2) is 1/6 arctan(x/6)+C

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integral of 1/(36+x^2)

\int \frac{1}{36 + x ^{2}} d x

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

Solution steps

\int \frac{1}{36 + x ^{2}} 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{x}{6}

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

Add a constant to the solution

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

\frac{27}{x ^{2} + 9}

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

x \to \infty lim (x - 6)

(8 cos (π x)), (8 x^{2} - 2), [- 0.5, 0.5]

\frac{\partial}{\partial y} ((x - 4) ln (x y))

What is the integral of 1/(36+x^2) ?
The integral of 1/(36+x^2) is 1/6 arctan(x/6)+C