Question

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