$\int \frac{d x}{32-2 x^2}=A \log (4-x)+B \log (4+x)+c$, then the value of $\mathrm{A}$ and $\mathrm{B}$ are…
$\int \frac{d x}{32-2 x^2}=A \log (4-x)+B \log (4+x)+c$, then the value of $\mathrm{A}$ and $\mathrm{B}$ are respectively (where $\mathrm{c}$ is a constant of integration)
$\frac{-1}{8}, \frac{1}{8}$
$\frac{1}{8}, \frac{-1}{8}$
$\frac{-1}{16}, \frac{1}{16}$
$\frac{1}{8}, \frac{1}{8}$
Solution
Let $I=\int \frac{d x}{32-2 x^2}$
$=\frac{1}{2}\left[\frac{1}{2(4)} \log \left|\frac{4+\mathrm{x}}{4-\mathrm{x}}\right|\right]+\mathrm{c}=\frac{1}{16}[\log |4+\mathrm{x}|-\log |4-\mathrm{x}|]+\mathrm{c}$
Comparing with given data we get $\mathrm{A}=\frac{-1}{16}, \mathrm{~B}=\frac{1}{16}$