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

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

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integral of 2/(9+x^2)

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

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

Solution steps

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

y = x + \frac{5}{x}

\frac{d}{d x} (- \frac{36}{x ^{4}})

\frac{d}{d x} (ln (3 x) - e^{- 2 x})

\int (cot (b x) + tan (b x))^{2} d x

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

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