Find the indefinite integral. (Note: Solve by the most convenient method-not all require integration by parts. Remember the constant of integration.) \( \int t \ln (t+2) d t \)

To solve the integral \( \int t \ln(t + 2) \, dt \), we can use integration by parts. Let: - \( u = \ln(t + 2) \) → \( du = \frac{1}{t + 2} \, dt \) - \( dv = t \, dt \) → \( v = \frac{t^2}{2} \) Now, applying integration by parts \( \int u \, dv = uv - \int v \, du \): \[ \int t \ln(t + 2) \, dt = \frac{t^2}{2} \ln(t + 2) - \int \frac{t^2}{2} \cdot \frac{1}{t + 2} \, dt \] The remaining integral can be simplified: \[ \int \frac{t^2}{2(t + 2)} \, dt = \frac{1}{2} \int \left( t - 2 + \frac{4}{t + 2} \right) \, dt \] Now, calculating the integrals separately: \[ = \frac{1}{2} \left( \frac{t^2}{2} - 2t + 4 \ln(t + 2) \right) + C \] Putting it all together, we obtain: \[ \int t \ln(t + 2) \, dt = \frac{t^2}{2} \ln(t + 2) - \frac{1}{4} t^2 + t - 2 \ln(t + 2) + C \] Thus, the final answer is: \[ \int t \ln(t + 2) \, dt = \frac{t^2}{2} \ln(t + 2) - \frac{1}{4} t^2 + t + C \]