What is the integral of 1/(x^2+9) ?

The integral of 1/(x^2+9) is 1/3 arctan(x/3)+C

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integral of 1/(x^2+9)

\int \frac{1}{x ^{2} + 9} d x

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

Solution steps

\int \frac{1}{x ^{2} + 9} d x

Apply Integral Substitution

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

Substitute back

u = \frac{x}{3}

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

Add a constant to the solution

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

f (x) = 2 x^{3} - x^{2} + 3 x, at x = 1

\int_{0}^{t} e^{x} \cdot sin (π \cdot x) d x

6 e^{x} cos (x)

x \to 6 lim (x^{2} - 36)

\frac{\partial}{\partial x} (5 x)

What is the integral of 1/(x^2+9) ?
The integral of 1/(x^2+9) is 1/3 arctan(x/3)+C