What is the integral of 1/(x^2+81) ?

The integral of 1/(x^2+81) is 1/9 arctan(x/9)+C

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integral of 1/(x^2+81)

\int \frac{1}{x ^{2} + 81} d x

\frac{1}{9} arctan (\frac{x}{9}) + C

Solution steps

\int \frac{1}{x ^{2} + 81} d x

Apply Integral Substitution

= \int \frac{1}{9 ( 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}{9} \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}{9} arctan (u)

Substitute back

u = \frac{x}{9}

= \frac{1}{9} arctan (\frac{x}{9})

Add a constant to the solution

= \frac{1}{9} arctan (\frac{x}{9}) + C

y = \frac{10}{ln ( x )}

x \to 0 lim (\frac{ln ( cos ( x ) )}{ln ( cos ( 2 x ) )})

\int (x - 3)^{3} d x

(x^{2} + 1) \frac{d y}{d x} + 5 x (y - 1) = 0, y (0) = 4

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

What is the integral of 1/(x^2+81) ?
The integral of 1/(x^2+81) is 1/9 arctan(x/9)+C