<s>[INST] How to evaluate $\lim_{c \rightarrow \infty} \int_{-c}^c f(x)dx$ I'm trying to evaluate:
$$\lim_{c \rightarrow \infty} \int_{-c}^c \frac{1+x}{1+x^2}dx$$
but I don't understand how to evaluate 
$$\lim_{c \rightarrow \infty} \int_{-c}^c f(x)dx$$
How?
 [/INST] Hint. Observe that
$$
\int_{-c}^c \frac{1+x}{1+x^2}dx=\int_{-c}^c \frac{1}{1+x^2}dx=2\int_0^c \frac{1}{1+x^2}dx=2\arctan(c).
$$ Can you take it from here?
Some details. We have
$$
\begin{align}
\color{blue}{\int_{-c}^c \frac{x}{1+x^2}dx}&=\int_{-c}^0 \frac{x}{1+x^2}dx+\int_0^c \frac{x}{1+x^2}dx
\\\\&=\int_{c}^0 \frac{-x}{1+(-x)^2}(-dx)+\int_0^c \frac{x}{1+x^2}dx
\\\\&=-\int_0^c \frac{x}{1+x^2}dx+\int_0^c \frac{x}{1+x^2}dx
\\\\&=\color{blue}{0}
\end{align}
$$ giving
$$
\int_{-c}^c \frac{1+x}{1+x^2}dx=\int_{-c}^c \frac{1}{1+x^2}dx+\color{blue}{\int_{-c}^c \frac{x}{1+x^2}dx}=\int_{-c}^c \frac{1}{1+x^2}dx+\color{blue}{0}.
$$
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