What is the integral of (3x-1)/(x^2+9) ?

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

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integral of (3x-1)/(x^2+9)

\int \frac{3 x - 1}{x ^{2} + 9} d x

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

Solution steps

\int \frac{3 x - 1}{x ^{2} + 9} d x

Expand

\frac{3 x - 1}{x ^{2} + 9} : \frac{3 x}{x ^{2} + 9} - \frac{1}{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{3 x}{x ^{2} + 9} d x - \int \frac{1}{x ^{2} + 9} d x

\int \frac{3 x}{x ^{2} + 9} d x = \frac{3}{2} ln x^{2} + 9

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

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

Add a constant to the solution

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

\frac{d}{d x} (ln (g (x)))

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

d er i v a t i v e \frac{3}{x - 1}

\int arctan (t^{2}) d t

t an g e n t f (x) = - 0.5 x^{2}, at x = 10

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