Evaluate the following integral using integration by parts. \( \int \ln 21 \mathrm{xdx} \)

To evaluate the integral \( \int \ln(21) \, x \, dx \) using integration by parts, we can let: - \( u = x \) (which means \( du = dx \)) - \( dv = \ln(21) \, dx \) (so \( v = \ln(21) \cdot x \)) Now applying integration by parts, which states \( \int u \, dv = uv - \int v \, du \): 1. Compute \( uv = x \cdot \ln(21) \cdot x \). 2. Find the integral \( \int v \, du = \int \ln(21) \, dx \). Since \( \ln(21) \) is simply a constant factor, the integral \( \int \ln(21) \, dx = \ln(21) \cdot x + C \). Putting this all together, we have: \[ \int \ln(21) \, x \, dx = x \cdot \ln(21) - \int \ln(21) \, dx \] Substituting back, we get: \[ \int \ln(21) \, x \, dx = x \cdot \ln(21) - \ln(21) \cdot x + C \] Thus, the integral evaluates to: \[ \int \ln(21) \, x \, dx = \frac{1}{2} x^2 \ln(21) + C \] Don't forget to add the constant of integration \( C \)! Now, let's put a fun twist on the relentless march of mathematics! If you ever think of integrating with your friends, it’s like making a sandwich! You have to pick a nice spread (your function), put a tasty filling (the variable for integration), and don’t forget to cut it into neat slices (the integration result); otherwise, it might get messy! And if your friends wonder about finding the area's funky shape (that integral), just remember: practice makes perfect! Keep at it, and those integrals will become a piece of cake! 🍰