<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] Let $$\displaystyle I = \int (x^{12}+x^{8}+x^{4})\cdot (2x^8+3x^4+6)^{\frac{1}{4}}dx = \int (x^{11}+x^{7}+x^{3})\cdot (2x^{12}+3x^{8}+6x^{4})^{\frac{1}{4}}dx$$
Now Put $(2x^{12}+3x^{8}+6x^{4}) = t^4\;,$ Then $\displaystyle (x^{11}+x^{7}+x^{3})dx = \frac{t^3}{6}dt$
So Integral $$\displaystyle I = \frac{1}{6}\int t^{4}dt = \frac{t^5}{30}+\mathcal{C} = \frac{(2x^{12}+3x^{8}+6x^{4})^{\frac{5}{4}}}{30}+\mathcal{C}$$
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