Question

Evaluate \[ \int x^2(\log x)^2\,dx. \]

• A. \[ \frac16x^3\left[3(\log x)^2-3\log x+4\right]+C \]
• B. \[ \frac1{27}x^3\left[9(\log x)^2-6\log x+2\right]+C \]
• C. \[ \frac1{27}x^3\left[9(\log x)^2+6\log x+2\right]+C \]
• D. \[ \frac19x^3\left[6(\log x)^2-3\log x+1\right]+C \]

Hint:

For integrals involving powers of \(\log x\), choose \((\log x)^n\) as the first function in integration by parts. Each application reduces the power of \(\log x\) by one.

Answer 1

The Correct Option is B

Concept: Use integration by parts repeatedly. \[ \int u\,dv=uv-\int v\,du. \]

Step 1: Apply integration by parts. Let \[ u=(\log x)^2, \qquad dv=x^2\,dx. \] Then \[ du=\frac{2\log x}{x}\,dx, \qquad v=\frac{x^3}{3}. \] Therefore, \[ I=\int x^2(\log x)^2\,dx = \frac{x^3}{3}(\log x)^2 -\frac23\int x^2\log x\,dx. \]

Step 2: Evaluate \(\int x^2\log x\,dx\). Let \[ u=\log x, \qquad dv=x^2\,dx. \] Then \[ du=\frac1x\,dx, \qquad v=\frac{x^3}{3}. \] Hence, \[ \int x^2\log x\,dx = \frac{x^3}{3}\log x -\frac13\int x^2\,dx. \] \[ = \frac{x^3}{3}\log x -\frac{x^3}{9}. \]

Step 3: Substitute back into \(I\). \[ I = \frac{x^3}{3}(\log x)^2 -\frac23 \left( \frac{x^3}{3}\log x -\frac{x^3}{9} \right). \] \[ = \frac{x^3}{3}(\log x)^2 -\frac{2x^3}{9}\log x +\frac{2x^3}{27}. \] Taking \(\dfrac{x^3}{27}\) common, \[ I = \frac{x^3}{27} \Big[ 9(\log x)^2 -6\log x +2 \Big] +C. \] Therefore, \[ \boxed{ \int x^2(\log x)^2\,dx = \frac1{27}x^3 \left[ 9(\log x)^2 -6\log x +2 \right] +C } \] \[ \boxed{\text{Answer = (B)}} \]