MCQ
The integral $\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{\left(2-x^{2}\right) d x}{\left(2+x^{2}\right) \sqrt{4+x^{4}}}$ is equal to
- A$1$
- B$2$
- C$4$
- ✓$3$
$\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{x^{2}\left(\frac{2}{x^{2}}-1\right) d x}{x\left(x+\frac{2}{x}\right) \times x \sqrt{\frac{4}{x^{2}}+x^{2}}}$
$\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{\left(\frac{2}{x^{2}}-1\right) d x}{\left(x+\frac{2}{x}\right) \sqrt{\left(x+\frac{2}{x}\right)^{2}-4}}$
$x +\frac{2}{ x }= t$
$dt =\left(1-\frac{2}{ x ^{2}}\right) dx$
$I =-\frac{24}{\pi} \int \frac{ dt }{ t \sqrt{ t ^{2}-4}}$
$=-\frac{24}{\pi} \times \frac{1}{2} \sec ^{-1}\left[\frac{ x +\frac{2}{ x }}{2}\right)^{\sqrt{2}}$
$=-\frac{12}{\pi}\left[\sec ^{-1}\left(\frac{2 \sqrt{2}}{2}\right)-\sec ^{-1}(\infty)\right]$
$=-\frac{12}{\pi}\left[\frac{\pi}{4}-\frac{2 \pi}{2 \times 2}\right]=-\frac{12}{\pi}\left[-\frac{\pi}{4}\right]$
$=3$
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$f(x)=\left\{\begin{array}{ll} \frac{\cos ^{-1}\left(1-\{x\}^{2}\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^{3}}, & x \neq 0 \\ \alpha, & x=0 \end{array}\right.$
is continuous at $x=0,$ where $\{x\}=x-[x],[x]$ is the greatest integer less than or equal to $X$.
Then :