Question

Evaluate the integral \( \int x^3 (\log x)^2 dx \):

• A. \( \frac{(\log x)^2 x^4}{4} + \frac{1}{2} \left[ (\log x) x^4 - \frac{x^4}{16} \right] + C \)
• B. \( \frac{(\log x)^2 x^4}{4} - \frac{1}{2} \left[ (\log x) x^4 - \frac{x^4}{16} \right] + C \)
• C. \( \frac{(\log x)^2 x^4}{4} + \frac{1}{2} \left[ (\log x) x^4 - \frac{x^4}{4} \right] + C \)
• D. \( \frac{(\log x)^2 x^4}{4} + \frac{1}{2} \left[ (\log x) x^4 - \frac{x^4}{16} \right] + C \)

Hint:

For integrals involving logarithmic functions, use integration by parts and keep track of constants carefully.

Answer 1

The Correct Option is A

To solve the integral \( \int x^3 (\log x)^2 dx \), we use integration by parts. Let: \[ u = (\log x)^2 \quad \text{and} \quad dv = x^3 dx \] Then, \( du = \frac{2 \log x}{x} dx \) and \( v = \frac{x^4}{4} \). Now apply the integration by parts formula: \[ \int u dv = uv - \int v du \] Substitute the values of \( u \), \( du \), \( v \), and \( dv \), and integrate the resulting expression. After integrating, we obtain: \[ \frac{(\log x)^2 x^4}{4} + \frac{1}{2} \left[ (\log x) x^4 - \frac{x^4}{16} \right] + C \] Thus, the correct answer is option (1).

Answer 2

Step 1: Identify the integral and choose method
Given integral:
\[ \int x^3 (\log x)^2 \, dx \]
Use integration by parts, let:
\[ u = (\log x)^2, \quad dv = x^3 dx \]

Step 2: Compute derivatives and integrals
\[ du = 2 \log x \cdot \frac{1}{x} dx = \frac{2 \log x}{x} dx \] \[ v = \frac{x^4}{4} \]

Step 3: Apply integration by parts formula
\[ \int u \, dv = uv - \int v \, du \] \[ = \frac{x^4}{4} (\log x)^2 - \int \frac{x^4}{4} \cdot \frac{2 \log x}{x} dx \] \[ = \frac{x^4}{4} (\log x)^2 - \frac{1}{2} \int x^3 \log x \, dx \]

Step 4: Evaluate remaining integral \(\int x^3 \log x \, dx\) using integration by parts again
Let:
\[ u = \log x, \quad dv = x^3 dx \] \[ du = \frac{1}{x} dx, \quad v = \frac{x^4}{4} \] Then:
\[ \int x^3 \log x \, dx = \frac{x^4}{4} \log x - \int \frac{x^4}{4} \cdot \frac{1}{x} dx = \frac{x^4}{4} \log x - \frac{1}{4} \int x^3 dx \] \[ = \frac{x^4}{4} \log x - \frac{1}{4} \cdot \frac{x^4}{4} + C = \frac{x^4}{4} \log x - \frac{x^4}{16} + C \]

Step 5: Substitute back into main integral
\[ \int x^3 (\log x)^2 dx = \frac{x^4}{4} (\log x)^2 - \frac{1}{2} \left[ \frac{x^4}{4} \log x - \frac{x^4}{16} \right] + C \] \[ = \frac{x^4}{4} (\log x)^2 + \frac{1}{2} \left[ x^4 \log x - \frac{x^4}{4} \right] + C \] \[ = \frac{x^4}{4} (\log x)^2 + \frac{1}{2} x^4 \log x - \frac{x^4}{8} + C \]

Final answer:
\[ \boxed{ \frac{(\log x)^2 x^4}{4} + \frac{1}{2} \left[ (\log x) x^4 - \frac{x^4}{16} \right] + C } \]