Evaluate the following integrals:

Let $\mathrm{I}=\int \frac{\mathrm{x}^{2}}{\sqrt{3 \mathrm{x}+4}} \mathrm{dx}$

Substituting $3 x+4=t \Rightarrow 3 d x=d t$

$\Rightarrow I=\int \frac{\left(\frac{t-4}{3}\right)^{2}}{3 \sqrt{t}} d t$

$\Rightarrow \mathrm{I}=\frac{1}{27} \int \frac{\mathrm{t}^{2}+16-8 \mathrm{t}}{\sqrt{\mathrm{t}}} \mathrm{dt}$

$\Rightarrow \mathrm{I}=\frac{1}{27} \int\left(\mathrm{t}^{\frac{3}{2}}-8 \mathrm{t}^{\frac{1}{2}}+16 \mathrm{t}^{-\frac{1}{2}}\right) \mathrm{dt}$

$\Rightarrow \mathrm{I}=\frac{1}{27}\left[\frac{2}{5} \mathrm{t}^{\frac{5}{2}}-\frac{16}{3} \mathrm{t}^{\frac{3}{2}}+32 \mathrm{t}^{\frac{1}{2}}\right]+\mathrm{c}$

$\Rightarrow I=\frac{1}{27}\left[\frac{2}{5}(3 x+4)^{\frac{5}{2}}-\frac{16}{3}(3 x+4)^{\frac{3}{2}}+32(3 x+4)^{\frac{1}{2}}\right]+c$

$\Rightarrow I=\frac{2}{135}(3 x+4)^{\frac{5}{2}}-\frac{16}{81}(3 x+4)^{\frac{3}{2}}+\frac{32}{27}(3 x+4)^{\frac{1}{2}}+c$

Therefore, $\int \frac{\mathrm{x}^{2}}{\sqrt{3 \mathrm{x}+4}} \mathrm{dx}$

$=\frac{2}{135}(3 \mathrm{x}+4)^{\frac{5}{2}}-\frac{16}{81}(3 \mathrm{x}+4)^{\frac{3}{2}}+\frac{32}{27}(3 \mathrm{x}+4)^{\frac{1}{2}}+\mathrm{c}$