Find the indefinite integral. (Note: Solve by the most convenient method-not all require integration by parts. Remember the constant of integration.) \( \int \frac{x e^{3 x}}{(3 x+1)^{2}} d x \)

To solve the integral \( \int \frac{x e^{3 x}}{(3 x+1)^{2}} d x \), we can use the substitution method. Let’s set \( u = 3x + 1 \), therefore \( du = 3dx \) or \( dx = \frac{du}{3} \). Additionally, we express \( x \) in terms of \( u \): \( x = \frac{u - 1}{3} \). Now we can rewrite the integral as follows: \[ \int \frac{\frac{u - 1}{3} e^{3\left(\frac{u-1}{3}\right)}}{u^{2}} \cdot \frac{du}{3} = \frac{1}{9} \int \frac{(u - 1)e^{u-1}}{u^{2}} du \] This splits into two simpler integrals: \[ \frac{1}{9} \left( \int \frac{u e^{u-1}}{u^{2}} du - \int \frac{e^{u-1}}{u^{2}} du \right) = \frac{1}{9} \left( \int \frac{e^{u-1}}{u} du - \int \frac{e^{u-1}}{u^{2}} du \right) \] Both of these integrals can be solved separately. After computing these integrals and back-substituting for \( u = 3x + 1 \), we will arrive at the final solution for the indefinite integral. Finally, don't forget to add the constant of integration \( C \) to your final answer.