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Q.
If $I_1=\int\limits_0^1 \frac{1+x^8}{1+x^4} d x$ and $I_2=\int\limits_0^1 \frac{1+x^9}{1+x^3} d x$, then :
Integrals
Solution:
$x \in(0,1)$
so $1+x^8>1+x^9$
and $1+x^3>1+x^4$
$\Rightarrow\left(1+x^8\right)\left(1+x^3\right)>\left(1+x^9\right)\left(x+x^4\right) $
$\Rightarrow \frac{1+x^8}{1+x^4}>\frac{1+x^9}{1+x^3} $
$\Rightarrow \int\limits_0^1 \frac{1+x^8}{1+x^4} d x>\int\limits_0^1 \frac{1+x^9}{1+x^3} d x \Rightarrow I_1>I_2$
Now again $\frac{1+x^8}{1+x^4} < 1 \because 1+x^4>1+x^8$
$\Rightarrow \int\limits_0^1 \frac{1+x^8}{1+x^4} dx < \int\limits_0^1 dx \Rightarrow I _1<1$