Question

The value of \(\int 16x^3 \log_e x dx\) is equal to

• A. \(4x^4 \log_e(x) + x^4 + C\)
• B. \(4x^4 \log_e(x) - x^4 + C\)
• C. \(4x^4 \log_e(x)x^3 + C\)
• D. \(4x^4 \log_e(x) - \frac{4}{3}x^3 + C\)
• E. \(4x^4 \log_e(x) - 4x^4 + C\)

Hint:

When integrating a log function multiplied by \(x^n\), the result is always in the form \(\frac{x^{n+1}}{n+1} \log x - \frac{x^{n+1}}{(n+1)^2}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To integrate a product of an algebraic and a logarithmic function, we use Integration by Parts (ILATE rule).

Step 2: Key Formula or Approach:
Formula: \(\int u dv = uv - \int v du\).
Take \(u = \log_e x\) and \(dv = 16x^3 dx\).

Step 3: Detailed Explanation:
Let \(u = \log_e x \implies du = \frac{1}{x} dx\).
Let \(dv = 16x^3 dx \implies v = \frac{16x^4}{4} = 4x^4\).
Applying the formula:
\[ I = (4x^4)(\log_e x) - \int (4x^4)(\frac{1}{x}) dx \]
\[ I = 4x^4 \log_e x - \int 4x^3 dx \]
\[ I = 4x^4 \log_e x - \frac{4x^4}{4} + C = 4x^4 \log_e x - x^4 + C \]

Step 4: Final Answer:
The result is \(4x^4 \log_e(x) - x^4 + C\).