What is the integral of (x+6)/(x^2+9) ?

The integral of (x+6)/(x^2+9) is 1/2 ln|x^2+9|+2arctan(x/3)+C

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integral of (x+6)/(x^2+9)

\int \frac{x + 6}{x ^{2} + 9} d x

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

Solution steps

\int \frac{x + 6}{x ^{2} + 9} d x

Expand

\frac{x + 6}{x ^{2} + 9} : \frac{x}{x ^{2} + 9} + \frac{6}{x ^{2} + 9}

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{x}{x ^{2} + 9} d x + \int \frac{6}{x ^{2} + 9} d x

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

\int \frac{6}{x ^{2} + 9} d x = 2 arctan (\frac{x}{3})

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

Add a constant to the solution

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

\int_{- 3}^{- 2} 5 (2 x + 4)^{\frac{1}{3}} d x

\frac{d y}{d x} = 4 - y

\int \frac{x ^{3}}{9} d x

f (x) = \frac{7 x}{3 + x ^{2}}

y^{^{'}} + \frac{y}{x} - y = 0, y (1) = 0

What is the integral of (x+6)/(x^2+9) ?
The integral of (x+6)/(x^2+9) is 1/2 ln|x^2+9|+2arctan(x/3)+C