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

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

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integral of (x+2)/(x^2+9)

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

\frac{1}{2} ln \frac{x ^{2}}{9} + 1 + \frac{2}{3} arctan (\frac{x}{3}) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{3 u + 2}{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{3 u + 2}{u ^{2} + 1} d u

Expand

\frac{3 u + 2}{u ^{2} + 1} : \frac{3 u}{u ^{2} + 1} + \frac{2}{u ^{2} + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int \frac{3 u}{u ^{2} + 1} d u + \int \frac{2}{u ^{2} + 1} d u)

\int \frac{3 u}{u ^{2} + 1} d u = \frac{3}{2} ln u^{2} + 1

\int \frac{2}{u ^{2} + 1} d u = 2 arctan (u)

= \frac{1}{3} (\frac{3}{2} ln u^{2} + 1 + 2 arctan (u))

Substitute back

u = \frac{x}{3}

= \frac{1}{3} (\frac{3}{2} ln (\frac{x}{3})^{2} + 1 + 2 arctan (\frac{x}{3}))

Simplify

\frac{1}{3} (\frac{3}{2} ln (\frac{x}{3})^{2} + 1 + 2 arctan (\frac{x}{3})) : \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + \frac{2}{3} arctan (\frac{x}{3})

= \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + \frac{2}{3} arctan (\frac{x}{3})

Add a constant to the solution

= \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + \frac{2}{3} arctan (\frac{x}{3}) + C

\frac{d}{d x} ((3 x^{2} - x^{2}) (2 x + 5))

\int \frac{1 - 2 x}{x} d x

y (1 + x^{2}) y^{^{'}} - x (2 + y^{2}) = 0, y (0) = 3

\int \frac{x ^{2}}{( 4 - x ^{3} ) ^{2}} d x

f (x) = sin (arccos (x))

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