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

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

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integral of 1/(x^2+36)

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

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

Solution steps

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

x \to - \infty lim (\frac{x ^{2} + 2}{x ^{3} + x ^{2} - 1})

x \to 0 lim (7 x^{3} + \frac{1}{4} x^{2})

\frac{\partial}{\partial y} (x \cdot ln (x + 2 y) + \frac{x}{y})

f (x) = \frac{1}{x + 2}

x \to 0 lim (2 x (1 - x))

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