I have to study the punctual and uniform domain of $$f_n(x)=\int_0^{x+n}\dfrac{du}{2e^u+\sin^2u}.$$
First I observed that the integrand function is defined in $\mathbb R$ and it is decreasing $\forall x\ge 0$.
For the punctual convergence $x\in\mathbb R$ is fixed and $n\to \infty$. Since $x+n\sim n\to\infty$ we have
$$\lim_{n\to\infty}\int_0^{x+n}\dfrac{du}{2e^u+\sin^2u}=\int_0^{+\infty}\dfrac{du}{2e^u+\sin^2u}\le\int_0^{+\infty}\dfrac{du}{2e^u}=\dfrac{1}{2}\implies$$
$$\forall x\in\mathbb R,\exists\lim_{n\to\infty}f_n(x).$$
So the domain of punctual convergence should be $D_p=\mathbb R$.
We know that a sequence of functions is uniformly convergent in $U$ iff $\underset{n\to\infty}\lim\underset{x\in U}\sup |f_n(x)-f(x)|=0$.
What I know about $f(x)$ is that $f(x)\le\dfrac{1}{2}$, so I'm not sure about how to proceed to determine the domain of uniform convergence.
PS: Consider $n<m<N$ (we still have to determine $N$) and $x\in \mathbb R^+$, with an arbitary positive $\varepsilon$ fixed. $$|f_m(x)-f_n(x)|=\Big |\int_0^{m+x} \dfrac{du}{2e^u+\sin^2u}-\int_0^{n+x}\dfrac{du}{2e^u+\sin^2u}\Big |=$$ $$=\int_{n+x}^{m+x} \dfrac{du}{2e^u+\sin^2u}\le \int_{n+x}^{m+x} \dfrac{du}{2e^u}=-\dfrac{e^{-u}}{2}\Big|_{n+x}^{m+x}=-\dfrac{e^{-(m+x)}}{2}+\dfrac{e^{-(n+x)}}{2}=\dfrac{1}{2e^x}\Big(\dfrac{1}{e^n}-\dfrac{1}{e^m}\Big)\le \dfrac{1}{2}\Big(\dfrac{1}{e^n}-\dfrac{1}{e^m}\Big).$$ For the fact that $n<m\implies \dfrac{1}{e^n}>\dfrac{1}{e^m}\implies 0<\dfrac{1}{e^n}-\dfrac{1}{e^m}<1$ we have $$ \dfrac{1}{2}\Big(\dfrac{1}{e^n}-\dfrac{1}{e^m}\Big)<\dfrac{1}{2}\implies \mathbb N\ni N>2\varepsilon$$
For $U=\mathbb{R^+}$ $$\underset{n\to\infty}\lim\underset{x\in U}\sup |f_n(x)-f(x)|=\underset{n\to\infty}\lim\underset{x\in U}\sup \int\limits_{x+n}^{+\infty}\dfrac{du}{2e^u+\sin^2u}\leqslant\\ \leqslant\lim\limits_{n \to \infty} \int\limits_{n}^{+\infty}\dfrac{du}{2e^u+\sin^2u}=0$$ But for $U=\mathbb{R}$ we have $\sup\limits_{x \in \mathbb{R}} \int\limits_{x+n}^{+\infty}\dfrac{du}{2e^u+\sin^2u}=\int\limits_{-\infty}^{+\infty}\dfrac{du}{2e^u+\sin^2u}$ where last diverges.