Mathematics

If $$ \int\left(x^{2010}+x^{\text {804 }}+x^{402}\right)\left(2 x^{160 8}+5 x^{402}+10\right)^{1 / 402} d x $$
$$ =\dfrac{1}{10 a}\left(2 x^{2010}+5 x^{804}+10 x^{402}\right)^{a / 402} . $$ Then $$ (a-400) $$ is equal to


SOLUTION
Let $$ I=\int\left(x^{2010}+x^{804}+x^{402}\right)\left(2 x^{1608}+5 x^{402}+10\right)^{1 / 402} d x $$
$$ =\int x\left(x^{2009}+x^{508}+x^{401}\right) \cdot\left(2 x^{1608}+5 x^{402}+10\right)^{1 / 402} d x $$
$$ =\int\left(x^{2009}+x^{803}+x^{401}\right) \cdot\left(2 x^{2010}+5 x^{304}+10^{402}\right)^{1 / 402} d x $$
Put $$ 2 x^{2010}+5 x^{804}+10^{x^{402}}=t $$
$$ \Rightarrow \quad 4020\left(x^{2009}+x^{803}+x^{401}\right) d x=d t $$
$$ \therefore \quad I=\int \dfrac{1}{4020} \cdot(t)^{1 / 402} d t=\dfrac{1}{4020} \cdot \dfrac{t^{1 / 420+1}}{1 / 402+1} $$
$$ \quad=\dfrac{1}{4020} \cdot \dfrac{t^{403 / 402}}{403 / 402} $$
$$ =\dfrac{1}{4030}\left(2 x^{2010}+5 x^{804}+10^{402}\right)^{403 / 402} $$
$$ \therefore a-400=3 $$
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Subjective Medium Published on 17th 09, 2020
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