Question

The integral \[ \int x^n (1 + \log x) \, dx \] is equal to: 

• A. \( x^n + C \)
• B. \( x^{2x} + C \)
• C. \( x^n \log x + C \)
• D. \( \frac{1}{2}(1 + \log x)^2 + C \)

Hint:

For integrals involving terms like \( x^n \log x \), use integration by parts. The result will often involve terms like \( x^n \log x \) and simpler powers of \( x \).

Answer 1

The Correct Option is A

Step 1: Identifying the form of the integral. We are tasked with evaluating the integral: \[ \int x^n (1 + \log x) \, dx \] This expression can be simplified by distributing the terms inside the parentheses: \[ = \int x^n \, dx + \int x^n \log x \, dx \] Step 2: Solving the first part of the integral. The first term is a simple power of \( x \), and its integral is straightforward: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C_1 \quad ({for } n \neq -1) \] Step 3: Solving the second part of the integral. To solve \( \int x^n \log x \, dx \), we use integration by parts: \[ \int u \, dv = uv - \int v \, du \] Let: - \( u = \log x \), so \( du = \frac{1}{x} \, dx \) - \( dv = x^n \, dx \), so \( v = \frac{x^{n+1}}{n+1} \) Applying the integration by parts formula: \[ \int x^n \log x \, dx = \frac{x^{n+1}}{n+1} \log x - \int \frac{x^{n+1}}{n+1} \cdot \frac{1}{x} \, dx \] Simplifying the second integral: \[ = \frac{x^{n+1}}{n+1} \log x - \frac{1}{n+1} \int x^n \, dx \] From Step 2, we already know that: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C_1 \] Thus, the second integral becomes: \[ = \frac{x^{n+1}}{n+1} \log x - \frac{x^{n+1}}{(n+1)^2} + C_2 \] Step 4: Final solution. Combining both parts of the integral: \[ \int x^n (1 + \log x) \, dx = \frac{x^{n+1}}{n+1} + \left( \frac{x^{n+1}}{n+1} \log x - \frac{x^{n+1}}{(n+1)^2} \right) + C \] Simplifying the expression: \[ = x^n + C \]