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

The integral of 1/(x(3x+1)) is ln|x|-ln|3x+1|+C

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

\int \frac{1}{x ( 3 x + 1 )} d x

ln ∣ x ∣ - ln ∣ 3 x + 1 ∣ + C

Solution steps

\int \frac{1}{x ( 3 x + 1 )} d x

Take the partial fraction of

\frac{1}{x ( 3 x + 1 )} : \frac{1}{x} - \frac{3}{3 x + 1}

= \int \frac{1}{x} - \frac{3}{3 x + 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{3}{3 x + 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - ln ∣ 3 x + 1 ∣

Add a constant to the solution

= ln ∣ x ∣ - ln ∣ 3 x + 1 ∣ + C

\frac{d}{d t} (sin^{4} (t))

(t^{2} - 20 t + 75) \frac{d y}{d t} = y

f (x) = 6 x^{2} e^{6 x} + 2 x e^{6 x}

\frac{\partial}{\partial x} (ln (x^{7} + y^{7}))

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

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