Ans: (a) $x(\ln x)^2 -2x \ln x +2x +C$ (b) $\pi[(\ln 2)^2-2\ln2+1]$
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$\begin{array}{cl}
& \dint (\ln x)^2 dx \\
= & x(\ln x)^2-\dint xd(\ln x)^2 \\
= & x(\ln x)^2-\dint x \times 2 \ln x \times \dfrac{1}{x} dx \\
= & x(\ln x)^2-2\dint \ln x dx \\
= & x(\ln x)^2-2x\ln x+2\dint xd\ln x \\
= & x(\ln x)^2-2x\ln x +2\dint x \times \dfrac{1}{x} dx \\
= & x(\ln x)^2-2x\ln x +2\dint dx \\
= & x(\ln x)^2-2x\ln x +2x +C
\end{array}$ - Note that when $y=0$, $x=0$. Therefore, the $x$-intercept of the graph is $0$.
Hence, the required volume
$\begin{array}{cl}
= & \pi\dint_0^1\left(\sqrt{x}\ln(x^2+1)\right)^2 dx \\
= & \pi \dint_0^1 x[\ln(x^2+1)]^2dx \\
= & \dfrac{\pi}{2} \dint_0^1 [\ln(x^2+1)]^2 d(x^2+1) \\
= & \dfrac{\pi}{2}\left[ (x^2+1)[\ln(x^2+1)]^2-2(x^2+1)\ln (x^2+1)+2(x^2+1)\right]_0^1 \\
= & \dfrac{\pi}{2} [2(\ln 2)^2-2(2)\ln 2+2(2)-(\ln 1)^2-2\ln 1-2] \\
= & \pi[(\ln 2)^2-2\ln2+1]
\end{array}$
