Is $k=1$ the only real number for which $\int_0^\infty\log ( k+\exp (-x^2)) dx$ converges? Some computations I did using wolfram alpha to compute  $\int_0^\infty\log ( k+\exp (-x^2)) dx$ varying the value of $k$ , I get that integral converges only for $k=1$ as shown here , Now I ask if what i have conjectred is true if:
$k=1$ the only real number for which $\int_0^\infty\log ( k+\exp(-x^2))) dx$ converges ?
And my conjecture is :

Conjecture:
  let $f$ be a real valued function , $k$ is a real number , if $\int_0^\infty\log ( k+f(x))) dx$ converges  then $k$ must be  equal $1$

 A: First note that the integrand is undefined for large $x$ if $k<0$. Hence we have to take $k \geq 0$. For $k=0$ the integral becomes $-\int_0^{\infty} x^{2}\, dx$ which is $-\infty$. Now let $k>0$. We can write the integral as $\int_0^{\infty} [\log\, k +\log (1+\frac 1 k e^{-x^{2}})]\, dx$ . Note that $\int_0^{\infty} \log (1+\frac 1 k e^{-x^{2}})\, dx<\infty$ (because $\log (1+t) \leq t$ for al $t>0$). So the given integral is finite iff the constant $\log\, k$ is integrable which is true iff the constant is $0$,i.e. $k=1$. 
The conjecture is false. Let $f(x)=e^{e^{-x^{2}}}$. Then the integral converges for $k=0$. 
Adding a constant $c$ to $f$ in this example you can make the integral converge for any specified value of $k$.
A: $$\log(k+e^{-x^2})=\log k+\log\left(1+\frac{e^{-x^2}}k\right)\sim\log k+\frac{e^{-x^2}}k.$$
Unless $k=1$, the first term makes the integral diverge.
$\color{blue}{k=2},\color{green}{k=1},\color{magenta}{k=\frac12}.$

A: We have that
$$k + \exp(-x^2) \geqslant k$$
So we must have $k \geqslant 0$. In that case
$$\log(k + \exp(-x^2)) \to \log(k)$$
Which means that we must have that $\log(k)=0$, i.e. $k=1$. Otherwise, we can esteem either from above (when $\log(k) < 0)$ or from below (when $\log(k) > 0$) it by a constant function and conclude that it's divergent. (It does not mean that it's convergent for $k=1$, just that it's divergent when $k \neq 1$; but Kavi proved that it's convergent for $k=1$).  
About your conjecture: If
$$\exists \int_0^{\infty}\log(k+f(x))\,\mathrm{d}x$$
and the integral is improper 'at $x =\infty$' only and
$$\exists \lim_{x \to \infty} f(x)=:K$$
then we must have that $K+k=1$, because in that case we have that
$$\log(k+f(x)) \to \log(k+K)$$
and $\log(k+K)=0$ is neccessary for the integrability.
