Integration

\hspace{-16}$Here I am Latexifying Aditiya Solution.......\\\\\\ Let $\bf{I = \int \left(x^{3m}+x^{2m}+x^{m}\right)\cdot \left(2x^{2m}+3x^{m}+6\right)^{\frac{1}{m}}dx}$\\\\\\ Let $\bf{x^m = t\;,}$ Then $\bf{mx^{m-1}dx = dt\Rightarrow dx = \frac{1}{m}\cdot \frac{x}{t}dt=\frac{1}{m}\cdot t^{\frac{1}{m}}\cdot \frac{1}{t}}$\\\\\\ So Integral is $\bf{I = \frac{1}{m}\int \left(t^3+t^2+t\right)\cdot \left(2t^2+3t+6\right)^{\frac{1}{m}}\cdot \frac{t^{\frac{1}{m}}}{t}dt}$\\\\\\ So $\bf{I = \frac{1}{m}\int \left(t^2+t+1\right)\cdot \left(2t^3+3t^2+t\right)^{\frac{1}{m}}dt}$\\\\\\ Now Let $\bf{\left(2t^3+3t^2+t\right)=u^m\;,}$ Then $\bf{(t^2+t+1)dt = \frac{1}{6}\cdot mu^{m-1}du}$\\\\\\ So $\bf{I = \frac{1}{6}\int u^{m}du = \frac{1}{6(m+1)}u^{m+1}+\mathbb{C}=\frac{1}{6(m+1)}\cdot \left(2t^3+3t^2+t\right)^{\frac{m+1}{m}}+\mathbb{C}}$\\\\\\

\hspace{-16}$So $\bf{I = \int \left(x^{3m}+x^{2m}+x^{m}\right)\cdot \left(2x^{2m}+3x^{m}+6\right)^{\frac{1}{m}}dx = \frac{1}{6(m+1)}\cdot \left(2x^{3m}+3x^{2m}+x^m\right)^{\frac{m+1}{m}}+\mathbb{C}}$\\\\\\