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

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

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

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

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

Solution steps

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

Apply Integral Substitution

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

Expand

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

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

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

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

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

Substitute back

u = \frac{x}{3}

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

(\frac{x}{3})^{2} = \frac{x ^{2}}{9}

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

Add a constant to the solution

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

\frac{\partial}{\partial x} ((x^{2} - 3 y) (x - 2))

\frac{\partial}{\partial y} (\frac{e ^{x}}{e ^{x} + e ^{y}})

\frac{\partial}{\partial x} (5 x^{3} y + 3 sin (x y))

\int (2^{x} - 2)^{2} d x

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

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