$\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$ $\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$
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 $$\frac{x^5}{30}(2x^8+3x^4+6)^{\frac54}+C$$
 A: $\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$=$\int x^4(x^8+x^4+1)(2x^8+3x^4+6)^{1/4}dx$
$\int (x^{11}+x^7+x^3)(2x^{12}+3x^{8}+6x^{4})^{1/4}dx$
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$$(2x^{8}+3x^{4}+6)^{5/4}$+cHope this helps..
A: Let $$\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}$$
A: We start by factoring out $x^4$:
$$
(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}=x^4(x^{8}+x^4+1)(2x^8+3x^4+6)^{1/4}.
$$
Next, we write
$$
x^{8}+x^4+1=\frac{1}{6}(6+3x^4+2x^8)+\frac{1}{2}x^4+\frac{2}{3}x^8,
$$
so the integrand can be written as
$$
\Bigl(\frac{1}{2}x^8+\frac{2}{3}x^{12}\Bigr)(2x^8+3x^4+6)^{1/4} +\frac{1}{6}x^4(2x^8+3x^4+6)^{5/4},
$$
or, factoring out $x^5/24$ in the first term,
$$
\frac{x^5}{24}\Bigl(12x^3+16x^{7}\Bigr)(2x^8+3x^4+6)^{1/4} +\frac{1}{6}x^4(2x^8+3x^4+6)^{5/4}.
$$
Hooray (this is really lucky!), this is a derivative of a product, since
$$
D(2x^8+3x^4+6)=16x^7+12x^3,
$$
we find that the expression above is
$$
\frac{x^5}{30}D\bigl[(2x^8+3x^4+6)^{5/4}\bigr]+\bigl[D(x^5/30)\bigr](2x^8+3x^4+6)^{5/4},
$$
i.e.
$$
D\Bigl[\frac{x^5}{30}(2x^8+3x^4+6)^{5/4}\bigr].
$$
Hence,
$$
\int (x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}\,dx=\frac{x^5}{30}(2x^8+3x^4+6)^{5/4}+c.
$$
