Evaluate $\int_{-3}^{3}\frac{dx}{3+f(x)}$ if $f(x)\cdot f(-x)=9$ 
Evaluate $\displaystyle\int_{-3}^{3}\dfrac{\mathrm dx}{3+f(x)}$ if $f(x)\cdot f(-x)=9$

Attempt: 
$I = \int_{-3}^{3}\dfrac{\mathrm dx}{3+f(x)}$
$I = \int_{-3}^3 \dfrac{\mathrm dx}{3+f(-x)}$
Then I added the two $I$s to see if it helps, but it didn't.
I even substituted $f(-x)= \dfrac{9}{f(x)} $ in second equation but that didn't help too. 
Please provide a hint on how to solve it. 
 A: \begin{align*}
\int_{-3}^3\frac{\mathrm dx}{3+f(x)}&=\int_{-3}^0\frac{\mathrm dx}{3+f(x)}+\int_0^3\frac{\mathrm dx}{3+f(x)}&&\\[4pt]
&=\int_{3}^0\frac{-\mathrm dt}{3+f(-t)}+\int_0^3\frac{\mathrm dx}{3+f(x)}&&\text{making }x=-t\text{ when }-3\le x\le0\\[4pt]
&=\int_0^3\frac{\mathrm dt}{3+\frac9{f(t)}}+\int_0^3\frac{\mathrm dx}{3+f(x)}&&\text{since }f(t)\cdot f(-t)=9\\[4pt]
&=\int_0^3\left(\frac1{3+9/f(x)}+\frac1{3+f(x)}\right)\mathrm dx&&\\[4pt]
&=\int_0^3\left(\frac{f(x)}{3\left[3+f(x)\right]}+\frac1{3+f(x)}\right)\mathrm dx&&\\[4pt]
&=\int_0^3\frac{f(x)+3}{3\left[3+f(x)\right]}\mathrm dx&&\\[4pt]
&=\frac13\int_0^3\mathrm dx&&\\[4pt]
&=1
\end{align*}
A: Like Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx$. OR Integral $\int_{1}^{2011} \frac{\sqrt{x}}{\sqrt{2012 - x} + \sqrt{x}}dx$, 
As $\displaystyle\int_a^bf(x)\,\mathrm dx=\int_a^bf(a+b-x)\,\mathrm dx$
So, if $\int_a^bf(x)\,\mathrm dx=I,$
$$2I=\int_a^b[f(x)+f(a+b-x)]\,\mathrm dx$$
Here 
$$2I=\int_{-3}^3\left(\dfrac1{3+f(x)}+\dfrac1{3+f(-x)}\right)\mathrm dx$$
$$=\int_{-3}^3\dfrac{3+f(x)+3+f(-x)}{9+3f(x)+3f(-x)+f(x)f(-x)}\mathrm dx$$
$$=\int_{-3}^3\dfrac{\mathrm dx}3$$
A: As $$\frac{f(x)+3}{f(x)+3}=\frac{\dfrac9{f(-x)}}{\dfrac9{f(-x)}+3}+\frac3{f(x)+3}=\frac3{f(-x)+3}+\frac3{f(x)+3},$$
by integration from $-3$ to $3$
$$6=6I.$$
