<s>[INST] $\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$ $\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$
I tried to solve this question but no luck.
My try:
$$\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx=\int x^4(x^8+x^4+1)(2x^8+3x^4+6)^{1/4}dx\\
\int x^4(x^8+x^4+1)x^2(2+3x^{-4}+6x^{-8})^{1/4}dx$$
Now i got stuck,please help me reach the answer.Answer is $$\frac{x^5}{30}(2x^8+3x^4+6)^{\frac54}+C$$
 [/INST] $\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$=$\int x^4(x^8+x^4+1)(2x^8+3x^4+6)^{1/4}dx$
$\int (x^{11}+x^7+x^3)(2x^{12}+3x^{8}+6x^{4})^{1/4}dx$
Now just put $(2x^{12}+3x^{8}+6x^{4})$=$t$ so that $dt$=24$(x^{11}+x^7+x^3)$$dx$ and hence your integral finally simplifies to
(1/24)$\int(t^{1/4})dx$ which can be simplified to (1/30)$t^{5/4}$+c
Now put back t and you get : $(1/30)$$(2x^{12}+3x^{8}+6x^{4})^{5/4}$+c=
$(1/30)x^5$$(2x^{8}+3x^{4}+6)^{5/4}$+cHope this helps..
</s>