Let  $I=\int \frac{3e^x+5\sin x+10\cos x}{e^x+4\sin x+3\cos x}dx=mx+n\ln (e^x+4\sin x+3\cos x)+C,$  where ‘C’ is constant of integration and  $m,n\in N$, then the value of   $(m^2+n^2)$  is

Since  $\int \frac{3e^2+5\sin x+10\cos x}{e^x+4\sin x+3\cos x}dx=mx+n\ln (e^x+4\sin x+3\cos x)+C$
Differentiable both sides with respect to ‘x’, we get 
 $\Rightarrow \frac{3e^x+5\sin x+10\cos x}{e^x+4\sin x+3\cos x}=m+\frac{n.(e^x+4\sin x-3\cos x)}{e^x+4\sin x+3\cos x}$
 $\Rightarrow m+n=3and4m-3n=5$
Solving m = 2 and n = 1
$Hence,(m^2+n^2)={\left(2\right)}^2+{\left(1\right)}^2=4+1=5$