TITLE: Evaluate $\int_{-5}^{+5}\frac{dx}{1+f(x)}$ from the given information
QUESTION [5 upvotes]: $f(-x)f'(x)-f(x)f'(-x)=0$ and $f(0)=1$. Then what is the value of $$\int\limits_{-5}^{+5}\frac{dx}{1+f(x)}$$

My attempt:
From the first piece of information,
$f(x)f(-x)=k$ (some constant). Differentiating both sides, $f(-x)f'(x)-f(x)f'(-x)=0$. Putting $x=0$, we get $k=1$ (from second info given).
Now,I can't proceed further.

REPLY [8 votes]: Since $f(x)=\frac1{f(-x)}$ we have
$$\int_{-5}^{5}\frac{dx}{1+f(x)}=\int_{-5}^{0}\frac{dx}{1+\frac1{f(-x)}}+\int_{0}^{5}\frac{dx}{1+f(x)}=\int_{0}^{5}\frac{dx}{1+\frac1{f(x)}}+\int_{0}^{5}\frac{dx}{1+f(x)}=$$
$$=\int_{0}^{5}\frac{f(x)}{1+f(x)}dx+\int_{0}^{5}\frac{dx}{1+f(x)}=\int_{0}^{5}\frac{1+f(x)}{1+f(x)}dx=\int_{0}^{5}dx=5$$

REPLY [3 votes]: Let $$I=\int_{-5}^5\frac{f(x)\,dx}{1+f(x)}\tag{1}$$ and make the substitution $x=-u$. Then $$I=\int_5^{-5}\frac{f(-x)\,dx}{1+f(-x)}=\int_{-5}^5\frac{\frac1{f(x)}\,dx}{1+\frac1{f(x)}}=\int_{-5}^5\frac{dx}{1+f(x)}\tag{2}$$ so adding $(1)$ and $(2)$ gives $$2I=\int_{-5}^5dx=10\implies\boxed{\int_{-5}^5\frac{dx}{1+f(x)}=5}$$

REPLY [2 votes]: Since $f(x)=\frac{1}{f(-x)}$,
$$\frac{1}{1+f(x)}=\frac{f(-x)}{f(-x)+1}$$
So for every $a>0$,
$$\int_{-a}^{a}\frac{dx}{1+f(x)}=\int_{-a}^{a}\frac{f(-x)dx}{f(-x)+1}=\int_{-a}^a\frac{f(x)dx}{1+f(x)}=2a-\int_{-a}^{a}\frac{dx}{1+f(x)}$$
whence the desired integral is equal to $a$, and for $a=5$, the result is $5$.