If $\int ({\mathrm{x}}^{2010}+{\mathrm{x}}^{804}+{\mathrm{x}}^{402}){(2{\mathrm{x}}^{1608}+5{\mathrm{x}}^{402}+10)}^{\frac{1}{402}}\mathrm{dx}=$$\frac{1}{4030}{(2{\mathrm{x}}^{2010}+5{\mathrm{x}}^{804}+10{\mathrm{x}}^{402})}^{\frac{\mathrm{a}}{402}}$ then sum of the digit in ‘a’ is 

$\begin{array}{l}\mathrm{t}=2{\mathrm{x}}^{2010}+5{\mathrm{x}}^{804}+10{\mathrm{x}}^{402}\\ \frac{\mathrm{dt}}{\mathrm{dx}}=4020{\mathrm{x}}^{2009}+4020{\mathrm{x}}^{803}+4020{\mathrm{x}}^{401}\\ \mathrm{dt}=4020\left({\mathrm{x}}^{2009}+{\mathrm{x}}^{803}+{\mathrm{x}}^{401}\right)\mathrm{dx}\\ \frac{\mathrm{dt}}{4020}=\left({\mathrm{x}}^{2009}+{\mathrm{x}}^{803}+{\mathrm{x}}^{401}\right)\mathrm{dx}\\ \int \left({\mathrm{x}}^{2010}+{\mathrm{x}}^{804}+{\mathrm{x}}^{402}\right){\left(2{\mathrm{x}}^{108}+5{\mathrm{x}}^{402}+10\right)}^{\frac{1}{402}}\mathrm{dx}\\ \int \mathrm{x}\left({\mathrm{x}}^{2009}+{\mathrm{x}}^{803}+{\mathrm{x}}^{401}\right){\left(2{\mathrm{x}}^{1608}+5{\mathrm{x}}^{402}+10\right)}^{\frac{1}{402}}\mathrm{dx}\end{array}$
$\begin{array}{l}\int \left({\mathrm{x}}^{2009}+{\mathrm{x}}^{803}+{\mathrm{x}}^{401}\right){\left(2{\mathrm{x}}^{2010}+5{\mathrm{x}}^{804}+10{\mathrm{x}}^{402}\right)}^{\frac{1}{402}}\mathrm{dx}\\ \int {\mathrm{t}}^{\frac{1}{402}}\frac{\mathrm{dt}}{4020}=\frac{1}{4020}\left[\frac{{\mathrm{t}}^{\frac{403}{402}}}{\frac{403}{402}}\right]=\frac{1}{4020}\times \frac{402}{403}{\mathrm{t}}^{\frac{403}{402}}\\ =\frac{1}{4030}{\left(2{\mathrm{x}}^{2010}+5{\mathrm{x}}^{804}+10{\mathrm{x}}^{402}\right)}^{\frac{403}{402}}\end{array}$