WEIRD INTEGRATION

Overrated! Well, I did by Dimensional analysis.

Biggest power of $x$ inside the integral is $2010+\dfrac{1608}{402}=2014$ . Now $dx$ also has the dimensions of $x$ . Therefore, its $2015$ on LHS.

Biggest power of $x$ in LHS is $2010×\dfrac{k}{402}=5k$

So, $5k=2015\Rightarrow k=403$ . Sum of digits $=4+0+3=7$

$\displaystyle I=\int (x^{2010}+x^{804}+x^{402})(2x^{1608}+5x^{402}+10)^{\frac{1}{402}} dx$

$\displaystyle = \int x(x^{2009}+x^{803}+x^{401})(2x^{1608}+5x^{402}+10)^{\frac{1}{402}} dx$

$\displaystyle = \int (x^{2009}+x^{803}+x^{401})(x^{402})^{\frac{1}{402}} (2x^{1608}+5x^{402}+10)^{\frac{1}{402}} dx$

$\displaystyle = \int (x^{2009}+x^{803}+x^{401})(2x^{2010}+5x^{804}+10x^{402})^{\frac{1}{402}} dx$

Set $u=2x^{2010}+5x^{804}+10x^{402} , du= 4020(x^{2009}+x^{803}+x^{401})$

$\displaystyle => I = \frac{1}{4020} \int u^{\frac{1}{402}} du = \frac{1}{4020} \frac{402}{403} u^{\frac{403}{402}} +C$

$\displaystyle = \frac{1}{4030} ( 2x^{2010}+5x^{804}+10x^{402})^{\frac{403}{402}} + C$

Therefore: $\boxed{k=403}$ .

differentiate on Both sides and compare like terms