Question

Evaluate the integral: \[ \int 16x^3 \log_e x \, dx \]

• A. \( \frac{4x^4}{\ln x} \)
• B. \( 16x^4 \ln x - 4x^4 \)
• C. \( 4x^4 \ln x - \frac{4x^4}{\ln x} \)
• D. \( 4x^4 \ln x + \frac{4x^4}{\ln x} \)

Hint:

When integrating functions that involve logarithms and polynomials, use integration by parts to simplify the expression. The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \).

Answer 1

The Correct Option is B

We are tasked with evaluating the integral: \[ I = \int 16x^3 \log_e x \, dx \] Step 1: Use integration by parts.
To solve this, we use the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] Let: \[ u = \log_e x \quad \text{and} \quad dv = 16x^3 dx \] Then: \[ du = \frac{1}{x} dx \quad \text{and} \quad v = \int 16x^3 dx = 4x^4 \]
Step 2: Apply the integration by parts formula.
Now, apply the formula: \[ I = uv - \int v \, du \] Substitute the values of \( u \), \( v \), \( du \), and \( dv \): \[ I = \log_e x \cdot 4x^4 - \int 4x^4 \cdot \frac{1}{x} dx \] Simplify the second integral: \[ I = 4x^4 \ln x - \int 4x^3 dx \] Now, integrate \( \int 4x^3 dx \): \[ I = 4x^4 \ln x - 4 \times \frac{x^4}{4} \] \[ I = 4x^4 \ln x - x^4 \] Thus, the value of the integral is: \[ \boxed{16x^4 \ln x - 4x^4} \]