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

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

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

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

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

Solution steps

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

Take the partial fraction of

\frac{1}{x ( 3 x + 1 ) ^{2}} : \frac{1}{x} - \frac{3}{3 x + 1} - \frac{3}{( 3 x + 1 ) ^{2}}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Simplify

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

Add a constant to the solution

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

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