If $[x]$ is the greatest integer $\leq x$, then the value of the integral $\int_{-0.9}^{0…
If $[x]$ is the greatest integer $\leq x$, then the value of the integral $\int_{-0.9}^{0.9}\left(\left[x^2\right]+\log \left(\frac{2-x}{2+x}\right)\right) d x$ is
$0.486$
$0.243$
$1.8$
0
Solution
$
\text { } \begin{aligned}
& \int_{-0.9}^{0.9}\left\{\left[x^2\right]+\log \left(\frac{2-x}{2+x}\right)\right\} d x \\
= & \int_{-0.9}^{0.9}\left[x^2\right] d x+\int_{-0.9}^{0.9} \log \left(\frac{2-x}{2+x}\right) d x
\end{aligned}
$
$
\begin{aligned}
& =0+\int_{-0.9}^{0.9} \log \left(\frac{2-x}{2+x}\right) d x \\
& \text { Put } x=-x \Rightarrow f(x)=\log \frac{2-x}{2+x} \\
& \text { and } f(-x)=\log \frac{2+x}{2-x} \\
& =-\log \frac{(2-x)}{2+x}=-f(x)
\end{aligned}
$
So, it is an odd function, hence Required integral $=0$