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

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

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

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

- \frac{x}{3 ( 3 x + 1 )} + \frac{1}{9} ln ∣ 3 x + 1 ∣ + C

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{1}{3 ( 3 x + 1 )} d x = - \frac{1}{9} ln ∣ 3 x + 1 ∣

= - \frac{x}{3 ( 3 x + 1 )} - (- \frac{1}{9} ln ∣ 3 x + 1 ∣)

Simplify

= - \frac{x}{3 ( 3 x + 1 )} + \frac{1}{9} ln ∣ 3 x + 1 ∣

Add a constant to the solution

= - \frac{x}{3 ( 3 x + 1 )} + \frac{1}{9} ln ∣ 3 x + 1 ∣ + C

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