Question

\(\displaystyle \int (\log x)^2 x^3\,dx=\dfrac{x^4}{32}f(x)+C \Rightarrow f(x)=\)

• A. \(8(\log x)^2-4\log x+1\)
• B. \(8\log x-4x^4+x^3\)
• C. \(8(\log x)^2+4x-x^2\)
• D. \(4(\log x)^2-4x^2+x+1\)

Hint:

For integrals of the form \[ \int x^n(\log x)^m\,dx, \] use integration by parts repeatedly by taking the logarithmic expression as the first function.

Answer 1

The Correct Option is A

Step 1: Write the given integral.
We have \[ \int x^3(\log x)^2\,dx \] Use integration by parts.
Let \[ u=(\log x)^2,\qquad dv=x^3\,dx \] Then, \[ du=\frac{2\log x}{x}\,dx \] and \[ v=\frac{x^4}{4} \]

Step 2: Apply integration by parts.
\[ \int x^3(\log x)^2\,dx = \frac{x^4}{4}(\log x)^2-\int \frac{x^4}{4}\cdot \frac{2\log x}{x}\,dx \] \[ = \frac{x^4}{4}(\log x)^2-\frac12\int x^3\log x\,dx \]

Step 3: Evaluate \(\displaystyle \int x^3\log x\,dx\).
Again using integration by parts, let \[ u=\log x,\qquad dv=x^3\,dx \] Then, \[ du=\frac{1}{x}\,dx \] and \[ v=\frac{x^4}{4} \] Therefore, \[ \int x^3\log x\,dx = \frac{x^4}{4}\log x-\int \frac{x^4}{4}\cdot \frac{1}{x}\,dx \] \[ = \frac{x^4}{4}\log x-\frac14\int x^3\,dx \] \[ = \frac{x^4}{4}\log x-\frac{x^4}{16} \]

Step 4: Substitute back.
\[ \int x^3(\log x)^2\,dx = \frac{x^4}{4}(\log x)^2-\frac12\left(\frac{x^4}{4}\log x-\frac{x^4}{16}\right) \] \[ = \frac{x^4}{4}(\log x)^2-\frac{x^4}{8}\log x+\frac{x^4}{32} \] Taking \(\dfrac{x^4}{32}\) common, \[ = \frac{x^4}{32}\left[8(\log x)^2-4\log x+1\right] \]

Step 5: Compare with the given form.
Given, \[ \int (\log x)^2x^3\,dx=\frac{x^4}{32}f(x)+C \] Therefore, \[ f(x)=8(\log x)^2-4\log x+1 \]

Step 6: Final conclusion.
Hence, \[ \boxed{8(\log x)^2-4\log x+1} \]