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

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

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integral of 2/(x^2+9)

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integral Substitution

= 2 \cdot \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

= 2 \cdot \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)

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

Substitute back

u = \frac{x}{3}

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

Simplify

2 \cdot \frac{1}{3} arctan (\frac{x}{3}) : \frac{2}{3} arctan (\frac{x}{3})

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

Add a constant to the solution

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

\frac{x d y}{d x} = 2 y

\frac{d y}{d x}, y = 2 9^{x}

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 2

x \to 1 lim (x^{2} + 1)

2 x y d y + (x^{2} + y^{2}) d x = 0

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