What is the integral of x/((3x+4)^2) ?

The integral of x/((3x+4)^2) is -x/(3(3x+4))+1/9 ln|3x+4|+C

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integral of x/((3x+4)^2)

\int \frac{x}{( 3 x + 4 ) ^{2}} d x

- \frac{x}{3 ( 3 x + 4 )} + \frac{1}{9} ln ∣ 3 x + 4 ∣ + C

Solution steps

\int \frac{x}{( 3 x + 4 ) ^{2}} d x

Apply Integration By Parts

: - \frac{x}{3 ( 3 x + 4 )} - \int - \frac{1}{3 ( 3 x + 4 )} d x

= - \frac{x}{3 ( 3 x + 4 )} - \int - \frac{1}{3 ( 3 x + 4 )} d x

\int - \frac{1}{3 ( 3 x + 4 )} d x = - \frac{1}{9} ln ∣ 3 x + 4 ∣

= - \frac{x}{3 ( 3 x + 4 )} - (- \frac{1}{9} ln ∣ 3 x + 4 ∣)

Simplify

= - \frac{x}{3 ( 3 x + 4 )} + \frac{1}{9} ln ∣ 3 x + 4 ∣

Add a constant to the solution

= - \frac{x}{3 ( 3 x + 4 )} + \frac{1}{9} ln ∣ 3 x + 4 ∣ + C

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What is the integral of x/((3x+4)^2) ?
The integral of x/((3x+4)^2) is -x/(3(3x+4))+1/9 ln|3x+4|+C