All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$\int \frac{x^{3}}{1+x^{4}}dx = \frac{1}{4} \mathrm{ln} (1 + x^{4}) + C$

(ii)
Given $u = x^{2}, dx = \frac{1}{2x} du$

$\int \frac{x}{1+x^{4}} dx$

$= \int \frac{x}{1 +u^{2}} \frac{1}{2x} du$

$= \frac{1}{2} \int \frac{1}{1+u^{2}} du$

$= \frac{1}{2} tan^{-1}u + C$

$= \frac{1}{2} tan^{-1} (x^{2}) + C$

(iii)
Using a Graphing Calculator, $\int_0^1 (\frac{x}{1+x^{4}})^{2} dx = 0.186$

Do remember to change the $dx$ when doing substitution. Students are reminded that the integration formula can be found in MF15. The last part can be solved using GC since there is no mention of exact value.