Let $f\left(x\right)$ and $g\left(x\right)$ are 2 differentiable functions satisfying

$f\left(x\right)+3g\left(x\right)=x^2+x+6$ 

$2f\left(x\right)+4g\left(x\right)=2x^2+4$

If $J=\underset{0}{\overset{\pi /4}{\int }}\ln \left(g\left({\tan }^{2}x\right)-f\left(\tan x\right)-8\right)dx$, then the value of $\frac{3\pi \ln 2}{J}$

$f\left(x\right)+3g\left(x\right)=x^2+x+6\mathrm{...}\left(\text{i}\right)$

$2f\left(x\right)+4g\left(x\right)=2x^2+4\mathrm{...}\left(\text{ii}\right)$

–          –              –         –    .

              $2g\left(x\right)=2x+8$

Hence, $g\left(x\right)=x+4$

$\therefore f\left(x\right)=x^2-2x-6$

Now $g\left({\tan }^{2}x\right)=4+{\tan }^{2}x$

and     $f\left(\tan x\right)={\tan }^{2}x-2\tan x-6$ 

$\left(g\left({\tan }^{2}x\right)-f\left(\tan x\right)-8\right)=\left(4+{\tan }^{2}x\right)-\left({\tan }^{2}x-2\tan x-6\right)-8=2\tan x+2$

$J=\underset{0}{\overset{\pi /4}{\int }}\ln \left(g\left({\tan }^{2}x\right)-f\left(\tan x\right)-8\right)dx=\underset{0}{\overset{\pi /4}{\int }}\ln \left(2\tan x+2\right)dx$

$=\ln 2\underset{0}{\overset{\pi /4}{\int }}dx+\underset{0}{\overset{\pi /4}{\int }}\ln \left(1+\tan x\right)dx$

$J=\frac{\pi }{4}\ln 2+\frac{\pi }{8}\ln 2=\frac{3\pi }{8}\ln 2$

$\therefore \frac{3\pi \ln 2}{J}=8$