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

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

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

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

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

Solution steps

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - \frac{2}{3} arctan (3 x)

Add a constant to the solution

= ln ∣ x ∣ - \frac{2}{3} arctan (3 x) + C

\frac{d}{d x} ((5 x + 1)^{3})

(9.1) (6.5)

\int \frac{3}{y} - \frac{1}{y + 3} d y

\frac{d}{d x} (e^{y} sin (x))

\int \frac{3}{x 1 + 2 x} d x

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