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Mathematics Question on integral

If I(x)=∫x2(log⁡x)2dxI(x)=\int x^{2}(\log x)^{2} d x and I(1)=0I( 1)=0, then I(x)I(x)

A

x318[8(log⁡x)2−3log⁡x]+718\frac{x^{3}}{18} \left[ 8\left(\log x\right)^{2} -3\log x\right] + \frac{7}{18}

B

x327[9(log⁡x)2+6log⁡x]−227\frac{x^{3}}{27} \left[9\left(\log x\right)^{2} +6 \log x\right] - \frac{2}{27}

C

x327[9(log⁡x)2−6log⁡x+2]−227\frac{x^{3}}{27} \left[9\left(\log x\right)^{2} - 6 \log x+2\right]- \frac{2}{27}

D

x327[9(log⁡x)2−6log⁡x+2]−227\frac{x^{3}}{27} \left[9 \left(\log x\right)^{2} -6 \log x +2 \right] - \frac{2}{27}

Answer

x327[9(log⁡x)2−6log⁡x+2]−227\frac{x^{3}}{27} \left[9\left(\log x\right)^{2} - 6 \log x+2\right]- \frac{2}{27}

Explanation

Solution

Given integral I(x)=∫x2(log⁡x)2dx=x33(log⁡x)2−∫x332log⁡xxdxI(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] =x33(log⁡x)2−23[x33(log⁡x)−∫x33(1x)dx]= \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] =x33(log⁡x)2−23[x33(log⁡x)−13x33]+C= \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 =x327[9(log⁡x)2−6(log⁡x)+2]+C=\frac{x^{3}}{27}\left[9(\log x)^{2}-6(\log x)+2\right]+C ∵I(1)=0\because I(1)=0 ∴227+C=0\therefore \frac{2}{27}+C=0 ⇒C=−227\Rightarrow C=-\frac{2}{27} ∴I(x)=x327[9(log⁡x)2−6(log⁡x)+2]−227\therefore I(x)=\frac{x^{3}}{27}\left[9(\log x)^{2}-6(\log x)+2\right]-\frac{2}{27}