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

Question

Aicharya · Accepted Answer

Given integral $\begin{aligned} & I(x)=\int x^2(\log x)^2 d x=\frac{x^3}{3}(\log x)^2-\int \frac{x^3}{3} \frac{2 \log x}{x} d x \ & \quad[\text {by integration by parts] } \ & =\frac{x^3}{3}(\log x)^2-\frac{2}{3}\left[\frac{x^3}{3}(\log x)-\int \frac{x^3}{3}\left(\frac{1}{x}\right) d x\right] \ & =\frac{x^3}{3}(\log x)^2-\frac{2}{3}\left[\frac{x^3}{3}(\log x)-\frac{1}{3} \frac{x^3}{3}\right]+C \ & \quad=\frac{x^3}{27}\left[9(\log x)^2-6(\log x)+2\right]+C \ & \because \quad I(1)=0 \ & \therefore \frac{2}{27}+C=0 \Rightarrow C=-\frac{2}{27} \ & \therefore I(x)=\frac{x^3}{27}\left[9(\log x)^2-6(\log x)+2\right]-\frac{2}{27} \end{aligned}$ Hence, option (3) is correct.

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

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