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

The integral of (3x+1)/(x^2+3x) is 1/3 ln|x|+8/3 ln|x+3|+C

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

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

\frac{1}{3} ln ∣ x ∣ + \frac{8}{3} ln ∣ x + 3 ∣ + C

Solution steps

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

Expand

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

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

\int \frac{1}{x ^{2} + 3 x} d x = \frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x + 3 ∣

= 3 ln ∣ x + 3 ∣ + \frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x + 3 ∣

Simplify

3 ln ∣ x + 3 ∣ + \frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x + 3 ∣ : \frac{1}{3} ln ∣ x ∣ + \frac{8}{3} ln ∣ x + 3 ∣

= \frac{1}{3} ln ∣ x ∣ + \frac{8}{3} ln ∣ x + 3 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ x ∣ + \frac{8}{3} ln ∣ x + 3 ∣ + C

\int \frac{1}{x ^{2} - 4 x} d x

\frac{e ^{- s}}{s ^{2} + 4}

\frac{d}{d x} (6 sin^{3} (x))

\frac{d y}{d x} = 4 y e^{x}

\frac{\partial}{\partial x} (ln (9 x^{2} + 2 y^{2} + 4))

What is the integral of (3x+1)/(x^2+3x) ?
The integral of (3x+1)/(x^2+3x) is 1/3 ln|x|+8/3 ln|x+3|+C