Question:
If $I=\int_{1}^{2} \frac{d x}{\sqrt{2 x^{3}-9 x^{2}+12 x+4}}$, then:
Correct Option: , 2
Solution:
$f(x)=\frac{1}{\sqrt{2 x^{3}-9 x^{2}+12 x+4}}$
$f^{\prime}(x)=\frac{-1}{2}\left(\frac{\left(6 x^{2}-18 x+12\right)}{\left(2 x^{3}-9 x^{2}-12 x+4\right)^{3 / 2}}\right)$
$=\frac{-6(x-1)(x-2)}{2\left(2 x^{3}-9 x^{2}+12 x+4\right)^{3 / 2}}$
$f(1)=\frac{1}{3}$ and $f(2)=\frac{1}{\sqrt{8}}$
It is increasing function
$\frac{1}{3} $\frac{1}{9}
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