Evaluate the following integral using integration by parts. \[ \int x^{11} \ln x d x \]

To evaluate the integral \( \int x^{11} \ln x \, dx \) using integration by parts, we set: - \( u = \ln x \) \(\Rightarrow du = \frac{1}{x} \, dx\) - \( dv = x^{11} \, dx \) \(\Rightarrow v = \frac{x^{12}}{12}\) Now, we apply the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] Substituting our chosen \(u\) and \(v\): \[ \int x^{11} \ln x \, dx = \ln x \cdot \frac{x^{12}}{12} - \int \frac{x^{12}}{12} \cdot \frac{1}{x} \, dx \] This simplifies to: \[ = \frac{x^{12} \ln x}{12} - \frac{1}{12} \int x^{11} \, dx \] Next, we evaluate the remaining integral: \[ \int x^{11} \, dx = \frac{x^{12}}{12} \] Thus, substituting back: \[ \int x^{11} \ln x \, dx = \frac{x^{12} \ln x}{12} - \frac{1}{12} \cdot \frac{x^{12}}{12} \] This gives: \[ = \frac{x^{12} \ln x}{12} - \frac{x^{12}}{144} \] Finally, we can combine the terms: \[ \int x^{11} \ln x \, dx = \frac{x^{12}}{12} \left( \ln x - \frac{1}{12} \right) + C \] where \(C\) is the constant of integration. Thus, the integral evaluates to: \[ \int x^{11} \ln x \, dx = \frac{x^{12}}{12} \ln x - \frac{x^{12}}{144} + C \]