$\int { (x^{3m}+x^{2m}+x^{m})(2x^{2m}+3x^{m}+6)^{1/m}}dx$ if $x>0$ For any natural number $m$,how to evaluate $\int { (x^{3m}+x^{2m}+x^{m})(2x^{2m}+3x^{m}+6)^{1/m}}dx$ if $x>0$ ?
 A: The integral is given by:
$$ C+\frac{x^{1+m}}{6+6m}\left(6+3x^m+2x^{2m}\right)^{1+\frac{1}{m}}$$
just replace $x^m$ with $u$ and integrate by parts.
A: Factor out $x$ from $x^{3m}+x^{2m}+x^m$ and multiply it with $2x^{2m}+3x^m+6$  i.e
$$\int \left(x^{3m}+x^{2m}+x^m\right)\left(2x^{2m}+3x^m+6\right)^{1/m}\,dx=\int\left(x^{3m-1}+x^{2m-1}+x^{m-1}\right)\left(2x^{3m}+3x^{2m}+6x^{m}\right)^{1/m}\,dx$$
Substitute $\left(2x^{3m}+3x^{2m}+6x^{m}\right)=u \Rightarrow \left(x^{3m-1}+x^{2m-1}+x^{m-1}\right)\,dx=\frac{du}{6m}$ i.e
$$\frac{1}{6m}\int u^{1/m} \,du=\frac{1}{6+6m} \left(2x^{3m}+3x^{2m}+6x^m\right)^{1+1/m}+C$$ 
A: Let $$\displaystyle I =  \int { (x^{3m}+x^{2m}+x^{m})(2x^{2m}+3x^{m}+6)^{1/m}}dx\;,$$ Where $x>0$
Now Let $x^{m}=t\;,$ Then $$\displaystyle mx^{m-1}dx = dt\Rightarrow dx = \frac{dt}{mx^{m-1}}=x\frac{dt}{mt} = \frac{t^{\frac{1}{m}}}{mt}dt$$
So $$\displaystyle I = \frac{1}{m}\int (t^3+t^2+t)\cdot (2t^2+3t+6)^{\frac{1}{m}}\cdot \frac{t^{\frac{1}{m}}}{t}dt$$
So $$\displaystyle I  = \frac{1}{m}\int (t^2+t+1)\cdot (2t^3+3t^2+6t)^{\frac{1}{m}}dt$$
So Let $(2t^3+3t^2+6t)=u\;,$ Then $\displaystyle (t^2+t+1)dt = \frac{du}{6}$
So $$\displaystyle I = \frac{1}{6m}\int u^{\frac{1}{m}}du = \frac{1}{6m}\cdot \frac{u^{\frac{1}{m}+1}}{\frac{1}{m}+1}+\mathcal{C} = \frac{1}{6}\cdot \frac{(2t^3+3t^2+6t)^{\frac{m+1}{m}}}{m+1}+\mathcal{C}$$
So $$\displaystyle I = \frac{1}{6}\cdot \frac{(2x^{3m}+3x^{2m}+6x^{m})^{\frac{m+1}{m}}}{m+1}+\mathcal{C}$$
