Question

If $I(x)=\int x^{2}(\log x)^{2} d x$ and $I( 1)=0$, then $I(x)$

• A. $\frac{x^{3}}{18} \left[ 8\left(\log x\right)^{2} -3\log x\right] + \frac{7}{18} $
• B. $\frac{x^{3}}{27} \left[9\left(\log x\right)^{2} +6 \log x\right] - \frac{2}{27} $
• C. $\frac{x^{3}}{27} \left[9\left(\log x\right)^{2} - 6 \log x+2\right]- \frac{2}{27} $
• D. $\frac{x^{3}}{27} \left[9 \left(\log x\right)^{2} -6 \log x +2 \right] - \frac{2}{27}$

Answer 1

Answer: (C)

Given integral
$I(x)=\int x^{2}(\log x)^{2} d x= \frac{x^{3}}{3}(\log x)^{2}-\int \frac{x^{3}}{3} \frac{2 \log x}{x} d x$
[by integration by parts]
$= \frac{x^{3}}{3}(\log x)^{2}-\frac{2}{3}\left[\frac{x^{3}}{3}(\log x)-\int \frac{x^{3}}{3}\left(\frac{1}{x}\right) d x\right]$
$= \frac{x^{3}}{3}(\log x)^{2}-\frac{2}{3}\left[\frac{x^{3}}{3}(\log x)-\frac{1}{3} \frac{x^{3}}{3}\right]+C$
$=\frac{x^{3}}{27}\left[9(\log x)^{2}-6(\log x)+2\right]+C$
$\because I(1)=0$
$\therefore \frac{2}{27}+C=0$
$\Rightarrow C=-\frac{2}{27}$
$\therefore I(x)=\frac{x^{3}}{27}\left[9(\log x)^{2}-6(\log x)+2\right]-\frac{2}{27}$