(a) Use

\[ x=\sqrt3\tan u. \]

Then

\[ dx=\sqrt3\sec^2u\,du. \]

Also,

\[ 3+x^2=3+3\tan^2u=3\sec^2u. \]

and

\[ x^3=(\sqrt3\tan u)^3=3\sqrt3\tan^3u. \]

Therefore

\[ \frac{x^3}{3+x^2}\,dx = \frac{3\sqrt3\tan^3u}{3\sec^2u}\cdot\sqrt3\sec^2u\,du. \]

\[ =3\tan^3u\,du. \]

When \(x=1\),

\[ 1=\sqrt3\tan u \]

so

\[ \tan u=\frac1{\sqrt3} \]

and

\[ u=\frac{\pi}{6}. \]

When \(x=3\),

\[ 3=\sqrt3\tan u \]

so

\[ \tan u=\sqrt3 \]

and

\[ u=\frac{\pi}{3}. \]

Hence

\[ I=\int_{\frac16\pi}^{\frac13\pi}3\tan^3u\,du. \]

(b) It is easier to divide first:

\[ \frac{x^3}{3+x^2}=x-\frac{3x}{x^2+3}. \]

So

\[ I=\int_1^3\left(x-\frac{3x}{x^2+3}\right)\,dx. \]

\[ I=\left[\frac{x^2}{2}-\frac32\ln(x^2+3)\right]_1^3. \]

\[ I=\left(\frac92-\frac32\ln12\right)-\left(\frac12-\frac32\ln4\right). \]

\[ I=4-\frac32\ln3. \]

Answer: \(4-\frac32\ln3\).